Evaluate $lim_{h rightarrow 0} {(x+h)^{99}-x^{99}over h}$












1












$begingroup$



$$lim_{h rightarrow 0} {(x+h)^{99}-x^{99}over h}$$




I need to factor this in order to get a limit.



I tried:



$$lim_{h rightarrow 0} {[(x+h)^{33}]^3-[(x)^{33}]^3over h}
\ lim_{h rightarrow 0} {[(x+h)^{33}-(x)^{33}][(x+h)^{66}+(x+h)^{33}(x)^{33}+(x)^{66}]over h}$$

However this does not seem helpful.



How do I approach this question?










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  • $begingroup$
    Do you mean $h to 0$?
    $endgroup$
    – MisterRiemann
    Nov 26 '18 at 21:52










  • $begingroup$
    yes, thank you @MisterRiemann
    $endgroup$
    – didgocks
    Nov 26 '18 at 21:53










  • $begingroup$
    Hint: Limit definition of the derivative of $f(x)=x^{99}$.
    $endgroup$
    – Anurag A
    Nov 26 '18 at 21:53












  • $begingroup$
    Use the binomial theorem.
    $endgroup$
    – gammatester
    Nov 26 '18 at 21:54






  • 1




    $begingroup$
    The decomposition you show is not particularly helpful. Just apply the binomial theorem, and show that most of the terms disappear as $h$ goes to $0.$ Or, rewrite it saying $a=x+h, lim_limits{ato x} frac {a^{99} - x^{99}}{a-x}$ then factor $a^{99} - x^{99} = (a-x)(a^{98} + a^{97}x + cdots + x^{98})$ cancel the common factor and let $a = x$
    $endgroup$
    – Doug M
    Nov 26 '18 at 21:57


















1












$begingroup$



$$lim_{h rightarrow 0} {(x+h)^{99}-x^{99}over h}$$




I need to factor this in order to get a limit.



I tried:



$$lim_{h rightarrow 0} {[(x+h)^{33}]^3-[(x)^{33}]^3over h}
\ lim_{h rightarrow 0} {[(x+h)^{33}-(x)^{33}][(x+h)^{66}+(x+h)^{33}(x)^{33}+(x)^{66}]over h}$$

However this does not seem helpful.



How do I approach this question?










share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you mean $h to 0$?
    $endgroup$
    – MisterRiemann
    Nov 26 '18 at 21:52










  • $begingroup$
    yes, thank you @MisterRiemann
    $endgroup$
    – didgocks
    Nov 26 '18 at 21:53










  • $begingroup$
    Hint: Limit definition of the derivative of $f(x)=x^{99}$.
    $endgroup$
    – Anurag A
    Nov 26 '18 at 21:53












  • $begingroup$
    Use the binomial theorem.
    $endgroup$
    – gammatester
    Nov 26 '18 at 21:54






  • 1




    $begingroup$
    The decomposition you show is not particularly helpful. Just apply the binomial theorem, and show that most of the terms disappear as $h$ goes to $0.$ Or, rewrite it saying $a=x+h, lim_limits{ato x} frac {a^{99} - x^{99}}{a-x}$ then factor $a^{99} - x^{99} = (a-x)(a^{98} + a^{97}x + cdots + x^{98})$ cancel the common factor and let $a = x$
    $endgroup$
    – Doug M
    Nov 26 '18 at 21:57
















1












1








1





$begingroup$



$$lim_{h rightarrow 0} {(x+h)^{99}-x^{99}over h}$$




I need to factor this in order to get a limit.



I tried:



$$lim_{h rightarrow 0} {[(x+h)^{33}]^3-[(x)^{33}]^3over h}
\ lim_{h rightarrow 0} {[(x+h)^{33}-(x)^{33}][(x+h)^{66}+(x+h)^{33}(x)^{33}+(x)^{66}]over h}$$

However this does not seem helpful.



How do I approach this question?










share|cite|improve this question











$endgroup$





$$lim_{h rightarrow 0} {(x+h)^{99}-x^{99}over h}$$




I need to factor this in order to get a limit.



I tried:



$$lim_{h rightarrow 0} {[(x+h)^{33}]^3-[(x)^{33}]^3over h}
\ lim_{h rightarrow 0} {[(x+h)^{33}-(x)^{33}][(x+h)^{66}+(x+h)^{33}(x)^{33}+(x)^{66}]over h}$$

However this does not seem helpful.



How do I approach this question?







algebra-precalculus limits functions






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 '18 at 4:44









Paramanand Singh

49.6k556163




49.6k556163










asked Nov 26 '18 at 21:51









didgocksdidgocks

6801823




6801823












  • $begingroup$
    Do you mean $h to 0$?
    $endgroup$
    – MisterRiemann
    Nov 26 '18 at 21:52










  • $begingroup$
    yes, thank you @MisterRiemann
    $endgroup$
    – didgocks
    Nov 26 '18 at 21:53










  • $begingroup$
    Hint: Limit definition of the derivative of $f(x)=x^{99}$.
    $endgroup$
    – Anurag A
    Nov 26 '18 at 21:53












  • $begingroup$
    Use the binomial theorem.
    $endgroup$
    – gammatester
    Nov 26 '18 at 21:54






  • 1




    $begingroup$
    The decomposition you show is not particularly helpful. Just apply the binomial theorem, and show that most of the terms disappear as $h$ goes to $0.$ Or, rewrite it saying $a=x+h, lim_limits{ato x} frac {a^{99} - x^{99}}{a-x}$ then factor $a^{99} - x^{99} = (a-x)(a^{98} + a^{97}x + cdots + x^{98})$ cancel the common factor and let $a = x$
    $endgroup$
    – Doug M
    Nov 26 '18 at 21:57




















  • $begingroup$
    Do you mean $h to 0$?
    $endgroup$
    – MisterRiemann
    Nov 26 '18 at 21:52










  • $begingroup$
    yes, thank you @MisterRiemann
    $endgroup$
    – didgocks
    Nov 26 '18 at 21:53










  • $begingroup$
    Hint: Limit definition of the derivative of $f(x)=x^{99}$.
    $endgroup$
    – Anurag A
    Nov 26 '18 at 21:53












  • $begingroup$
    Use the binomial theorem.
    $endgroup$
    – gammatester
    Nov 26 '18 at 21:54






  • 1




    $begingroup$
    The decomposition you show is not particularly helpful. Just apply the binomial theorem, and show that most of the terms disappear as $h$ goes to $0.$ Or, rewrite it saying $a=x+h, lim_limits{ato x} frac {a^{99} - x^{99}}{a-x}$ then factor $a^{99} - x^{99} = (a-x)(a^{98} + a^{97}x + cdots + x^{98})$ cancel the common factor and let $a = x$
    $endgroup$
    – Doug M
    Nov 26 '18 at 21:57


















$begingroup$
Do you mean $h to 0$?
$endgroup$
– MisterRiemann
Nov 26 '18 at 21:52




$begingroup$
Do you mean $h to 0$?
$endgroup$
– MisterRiemann
Nov 26 '18 at 21:52












$begingroup$
yes, thank you @MisterRiemann
$endgroup$
– didgocks
Nov 26 '18 at 21:53




$begingroup$
yes, thank you @MisterRiemann
$endgroup$
– didgocks
Nov 26 '18 at 21:53












$begingroup$
Hint: Limit definition of the derivative of $f(x)=x^{99}$.
$endgroup$
– Anurag A
Nov 26 '18 at 21:53






$begingroup$
Hint: Limit definition of the derivative of $f(x)=x^{99}$.
$endgroup$
– Anurag A
Nov 26 '18 at 21:53














$begingroup$
Use the binomial theorem.
$endgroup$
– gammatester
Nov 26 '18 at 21:54




$begingroup$
Use the binomial theorem.
$endgroup$
– gammatester
Nov 26 '18 at 21:54




1




1




$begingroup$
The decomposition you show is not particularly helpful. Just apply the binomial theorem, and show that most of the terms disappear as $h$ goes to $0.$ Or, rewrite it saying $a=x+h, lim_limits{ato x} frac {a^{99} - x^{99}}{a-x}$ then factor $a^{99} - x^{99} = (a-x)(a^{98} + a^{97}x + cdots + x^{98})$ cancel the common factor and let $a = x$
$endgroup$
– Doug M
Nov 26 '18 at 21:57






$begingroup$
The decomposition you show is not particularly helpful. Just apply the binomial theorem, and show that most of the terms disappear as $h$ goes to $0.$ Or, rewrite it saying $a=x+h, lim_limits{ato x} frac {a^{99} - x^{99}}{a-x}$ then factor $a^{99} - x^{99} = (a-x)(a^{98} + a^{97}x + cdots + x^{98})$ cancel the common factor and let $a = x$
$endgroup$
– Doug M
Nov 26 '18 at 21:57












5 Answers
5






active

oldest

votes


















3












$begingroup$

If you're familiar with derivatives, then you can recall that
$$ f'(x) = lim_{hto 0} frac{f(x+h)-f(x)}{h} $$
and consider $f(x) = x^{99}$.



Otherwise, use the binomial theorem:
begin{align}frac{(x+h)^{99}-x^{99}}{h} &= frac{1}{h}left( sum_{k=0}^{99} binom{99}{k}x^{99-k}h^{k} - x^{99} right) = frac{1}{h}sum_{k=1}^{99}binom{99}{k}x^{99-k}h^{k}\&= binom{99}{1}x^{98} + sum_{k=2}^{99}binom{99}{k}x^{99-k}h^{k-1}
to 99x^{98}, quad text{as }hto 0.
end{align}






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$endgroup$









  • 1




    $begingroup$
    You made a typing mistake: It should be $f(x+h)-f(x)$, not $f(x+h)-f(h)$.
    $endgroup$
    – Martin Rosenau
    Nov 26 '18 at 21:58










  • $begingroup$
    @MartinRosenau Thanks!
    $endgroup$
    – MisterRiemann
    Nov 26 '18 at 21:58



















1












$begingroup$

Hint: That’s basically the derivative of $f(x) = x^{99}$.



$$lim_{h rightarrow 0} {(x+h)^{99}-x^{99}over h}$$



Recall that by Binomial Expansion,



$$(a+b)^n = sum_{r = 0}^{n} {n choose r}a^{n-r}b^r$$



So you get



$$lim_{h rightarrow 0} {color{blue}{{99 choose 0}x^{99}}+{99 choose 1}x^{98}h+{99 choose 2}x^{97}h^2+…+{99 choose n}h^{99}color{red}{-x^{99}}over h}$$



Cancel out the first and last terms. From here, notice if anything can be factored. Also, as another hint (for later simplifications), $${n choose r} = frac{n!}{r!{(n-r)!}}$$






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$endgroup$





















    1












    $begingroup$

    The trick is to realise that $h to 0$ means $h$ is arbitrarily small so we can assume $h < 1$ and so $h > h^2 > h^3$ and that the "higher values of $h$" become "negligible"



    So you are thinking way too hard. Just use the binomial theorem to get $(x + h)^{99} = x^{99} + 99hx^{98} + sum_{k=2}^{99} {99 choose k} h^kx^{99-k}$ and hope that must of those higher values will become negligible.



    And indeed:



    $require{cancel}limlimits_{hto 0}frac {(x + h)^{99}-x^{99}}{h }=$



    $limlimits_{hto 0}frac {x^{99} + 99hx^{98} + (sum_{k=2}^{99} {99 choose k} h^kx^{99-k})-x^{99}}{h}=$



    $limlimits_{hto 0}frac {color{green}{cancel{x^{99}}} + 99color{red}{cancel h}x^{98} + (sum_{k=2}^{99} {99 choose k} h^{color{red}{cancel {k}}k-1}x^{99-k}) - color{green}{cancel{x^{99}}}}{color{red}{cancel h}}=$



    $limlimits_{hto 0}(99x^{98} + sum_{k=2}^{99} {99 choose k}h^{k-1}x^{99-k})=$



    $99x^{98} + sum_{k=2}^{99}{99 choose k}(limlimits_{hto 0} h^{k-1})x^{99-k}=$



    $99x^{98} + sum_{k=2}^{99}{99 choose k}(0)x^{99-k}=$



    $99x^{98} + sum_{k=2}^{99}0=$



    $99x^{98}$



    ======



    Doug M had an intriguing hint in the comments:



    Considering $a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b+ .... + ab^{n-2}+ b^{99})$ we get:



    $(x + h)^{99} - x^{99}=$



    $[(x+h) - x][(x+h)^{98} + (x+h)^{97}x + ... + (x+h)x^{97} + x^{98}]=$



    $h[(x+h)^{98} + (x+h)^{97}x + ... + (x+h)x^{97} + x^{98}]$



    And



    $[(x+h)^{98} + (x+h)^{97}x + ... + (x+h)x^{97} + x^{98}]=$



    $(x^{98} + text{a bunch of stuff with h as a factor})+(x^{97}cdot x + x(text{a bunch of stuff with h as a factor}))+... (xcdot x^{97} + x^{97}(text{a bunch of stuff with h as a factor})) + x^{98})=$



    $99x^{98} + text { a REALLY big bunch of stuff with h as a factor}$



    So $frac {(x+h)^{98} - x^{99}}h=$



    $frac {h(99x^{98}+ text { a REALLY big bunch of stuff with h as a factor}}h = $



    $99x^{98} + text { a REALLY big bunch of stuff with h as a factor}$



    ... and "$ text { a REALLY big bunch of stuff with h as a factor}$" goes to $0$ as .... it's a REALLY big bunch of stuff with $h$ as a factor.



    Okay, .... it was an intriguing idea and I'm glad I did it but.... I don't think it's a very practical way to do it.






    share|cite|improve this answer











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    • 1




      $begingroup$
      Hoping that you don't mind, I shall adopt and reuse the REALLY big bunch of stuff
      $endgroup$
      – Claude Leibovici
      Nov 27 '18 at 6:01



















    0












    $begingroup$

    Because both $limlimits_{hto 0}((x-h)^{99}-x^{99})=0$ and $limlimits_{hto 0}h=0$ you may also use L'Hospital's rule:



    $displaystylelimlimits_{hto 0}frac{(x-h)^{99}-x^{99}}{h}=displaystylelimlimits_{hto 0}frac{frac{d}{dh}((x-h)^{99}-x^{99})}{frac{d}{dh}h}=displaystylelimlimits_{hto 0}frac{99(x-h)^{98}-0}{1}=99x^{98}$






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      L'Hospital is far too complicated. If you know derivatives, you can use the definition in this case.
      $endgroup$
      – gammatester
      Nov 26 '18 at 22:05



















    0












    $begingroup$

    This limit is the derivative of $f(x)=x^{99}$. To get a generalization you can replace $99$ with $n$ where $n in mathbb{R}$. The limit thus translates to $displaystyle lim_{h to 0} frac{(x+h)^{n}-x^n}{h}$.



    Using Binomial Theorem, which states that $(a+b)^n=displaystyle sum_{k=0}^n {n choose k} a^{n-k} b^k$. Substituting $begin{pmatrix}a \ bend{pmatrix}=begin{pmatrix}x \ hend{pmatrix}$. $$(x+h)^n=x^n+{n choose 1}x^{n-1} h + {n choose 2} x^{n-2} h^2+ cdots +{n choose n-1}xh^{n-1}+h^n$$ Putting back this result in the expression for the limit. We get the following limit $$displaystyle lim_{h to 0} frac{(x+h)^n-x^n}{h}=displaystyle lim_{h to 0} Biggl(frac{1}{h}{n choose 1}x^{n-1}h+frac{1}{h}biggl(h^2biggl({n choose 2}x^{n-2}+{n choose 3}x^{n-3}h+cdots+h^{n-2}biggr)biggr)Biggr)$$ $$=displaystyle lim_{h to 0} Biggl(biggl(nx^{n-1}biggr)+hbiggl({n choose 2}x^{n-2}+{n choose 3}x^{n-3}h+cdots+h^{n-2}biggr)Biggr)$$ $$=displaystyle lim_{h to 0}nx^{n-1}+underbrace{displaystyle lim_{h to 0}hbiggl({n choose 2}x^{n-2}+{n choose 3}x^{n-3}h+cdots+h^{n-2}biggr)}_{= 0}$$
    $$implies displaystyle lim_{h to 0}frac{(x+h)^n-x^n}{h}= nx^{n-1}$$
    Therefore, For $n=99$, the given limit evaluates to $99x^{98}$. Cheers






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      5 Answers
      5






      active

      oldest

      votes








      5 Answers
      5






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

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      3












      $begingroup$

      If you're familiar with derivatives, then you can recall that
      $$ f'(x) = lim_{hto 0} frac{f(x+h)-f(x)}{h} $$
      and consider $f(x) = x^{99}$.



      Otherwise, use the binomial theorem:
      begin{align}frac{(x+h)^{99}-x^{99}}{h} &= frac{1}{h}left( sum_{k=0}^{99} binom{99}{k}x^{99-k}h^{k} - x^{99} right) = frac{1}{h}sum_{k=1}^{99}binom{99}{k}x^{99-k}h^{k}\&= binom{99}{1}x^{98} + sum_{k=2}^{99}binom{99}{k}x^{99-k}h^{k-1}
      to 99x^{98}, quad text{as }hto 0.
      end{align}






      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        You made a typing mistake: It should be $f(x+h)-f(x)$, not $f(x+h)-f(h)$.
        $endgroup$
        – Martin Rosenau
        Nov 26 '18 at 21:58










      • $begingroup$
        @MartinRosenau Thanks!
        $endgroup$
        – MisterRiemann
        Nov 26 '18 at 21:58
















      3












      $begingroup$

      If you're familiar with derivatives, then you can recall that
      $$ f'(x) = lim_{hto 0} frac{f(x+h)-f(x)}{h} $$
      and consider $f(x) = x^{99}$.



      Otherwise, use the binomial theorem:
      begin{align}frac{(x+h)^{99}-x^{99}}{h} &= frac{1}{h}left( sum_{k=0}^{99} binom{99}{k}x^{99-k}h^{k} - x^{99} right) = frac{1}{h}sum_{k=1}^{99}binom{99}{k}x^{99-k}h^{k}\&= binom{99}{1}x^{98} + sum_{k=2}^{99}binom{99}{k}x^{99-k}h^{k-1}
      to 99x^{98}, quad text{as }hto 0.
      end{align}






      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        You made a typing mistake: It should be $f(x+h)-f(x)$, not $f(x+h)-f(h)$.
        $endgroup$
        – Martin Rosenau
        Nov 26 '18 at 21:58










      • $begingroup$
        @MartinRosenau Thanks!
        $endgroup$
        – MisterRiemann
        Nov 26 '18 at 21:58














      3












      3








      3





      $begingroup$

      If you're familiar with derivatives, then you can recall that
      $$ f'(x) = lim_{hto 0} frac{f(x+h)-f(x)}{h} $$
      and consider $f(x) = x^{99}$.



      Otherwise, use the binomial theorem:
      begin{align}frac{(x+h)^{99}-x^{99}}{h} &= frac{1}{h}left( sum_{k=0}^{99} binom{99}{k}x^{99-k}h^{k} - x^{99} right) = frac{1}{h}sum_{k=1}^{99}binom{99}{k}x^{99-k}h^{k}\&= binom{99}{1}x^{98} + sum_{k=2}^{99}binom{99}{k}x^{99-k}h^{k-1}
      to 99x^{98}, quad text{as }hto 0.
      end{align}






      share|cite|improve this answer











      $endgroup$



      If you're familiar with derivatives, then you can recall that
      $$ f'(x) = lim_{hto 0} frac{f(x+h)-f(x)}{h} $$
      and consider $f(x) = x^{99}$.



      Otherwise, use the binomial theorem:
      begin{align}frac{(x+h)^{99}-x^{99}}{h} &= frac{1}{h}left( sum_{k=0}^{99} binom{99}{k}x^{99-k}h^{k} - x^{99} right) = frac{1}{h}sum_{k=1}^{99}binom{99}{k}x^{99-k}h^{k}\&= binom{99}{1}x^{98} + sum_{k=2}^{99}binom{99}{k}x^{99-k}h^{k-1}
      to 99x^{98}, quad text{as }hto 0.
      end{align}







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Nov 26 '18 at 21:59

























      answered Nov 26 '18 at 21:57









      MisterRiemannMisterRiemann

      5,8451624




      5,8451624








      • 1




        $begingroup$
        You made a typing mistake: It should be $f(x+h)-f(x)$, not $f(x+h)-f(h)$.
        $endgroup$
        – Martin Rosenau
        Nov 26 '18 at 21:58










      • $begingroup$
        @MartinRosenau Thanks!
        $endgroup$
        – MisterRiemann
        Nov 26 '18 at 21:58














      • 1




        $begingroup$
        You made a typing mistake: It should be $f(x+h)-f(x)$, not $f(x+h)-f(h)$.
        $endgroup$
        – Martin Rosenau
        Nov 26 '18 at 21:58










      • $begingroup$
        @MartinRosenau Thanks!
        $endgroup$
        – MisterRiemann
        Nov 26 '18 at 21:58








      1




      1




      $begingroup$
      You made a typing mistake: It should be $f(x+h)-f(x)$, not $f(x+h)-f(h)$.
      $endgroup$
      – Martin Rosenau
      Nov 26 '18 at 21:58




      $begingroup$
      You made a typing mistake: It should be $f(x+h)-f(x)$, not $f(x+h)-f(h)$.
      $endgroup$
      – Martin Rosenau
      Nov 26 '18 at 21:58












      $begingroup$
      @MartinRosenau Thanks!
      $endgroup$
      – MisterRiemann
      Nov 26 '18 at 21:58




      $begingroup$
      @MartinRosenau Thanks!
      $endgroup$
      – MisterRiemann
      Nov 26 '18 at 21:58











      1












      $begingroup$

      Hint: That’s basically the derivative of $f(x) = x^{99}$.



      $$lim_{h rightarrow 0} {(x+h)^{99}-x^{99}over h}$$



      Recall that by Binomial Expansion,



      $$(a+b)^n = sum_{r = 0}^{n} {n choose r}a^{n-r}b^r$$



      So you get



      $$lim_{h rightarrow 0} {color{blue}{{99 choose 0}x^{99}}+{99 choose 1}x^{98}h+{99 choose 2}x^{97}h^2+…+{99 choose n}h^{99}color{red}{-x^{99}}over h}$$



      Cancel out the first and last terms. From here, notice if anything can be factored. Also, as another hint (for later simplifications), $${n choose r} = frac{n!}{r!{(n-r)!}}$$






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        Hint: That’s basically the derivative of $f(x) = x^{99}$.



        $$lim_{h rightarrow 0} {(x+h)^{99}-x^{99}over h}$$



        Recall that by Binomial Expansion,



        $$(a+b)^n = sum_{r = 0}^{n} {n choose r}a^{n-r}b^r$$



        So you get



        $$lim_{h rightarrow 0} {color{blue}{{99 choose 0}x^{99}}+{99 choose 1}x^{98}h+{99 choose 2}x^{97}h^2+…+{99 choose n}h^{99}color{red}{-x^{99}}over h}$$



        Cancel out the first and last terms. From here, notice if anything can be factored. Also, as another hint (for later simplifications), $${n choose r} = frac{n!}{r!{(n-r)!}}$$






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          Hint: That’s basically the derivative of $f(x) = x^{99}$.



          $$lim_{h rightarrow 0} {(x+h)^{99}-x^{99}over h}$$



          Recall that by Binomial Expansion,



          $$(a+b)^n = sum_{r = 0}^{n} {n choose r}a^{n-r}b^r$$



          So you get



          $$lim_{h rightarrow 0} {color{blue}{{99 choose 0}x^{99}}+{99 choose 1}x^{98}h+{99 choose 2}x^{97}h^2+…+{99 choose n}h^{99}color{red}{-x^{99}}over h}$$



          Cancel out the first and last terms. From here, notice if anything can be factored. Also, as another hint (for later simplifications), $${n choose r} = frac{n!}{r!{(n-r)!}}$$






          share|cite|improve this answer











          $endgroup$



          Hint: That’s basically the derivative of $f(x) = x^{99}$.



          $$lim_{h rightarrow 0} {(x+h)^{99}-x^{99}over h}$$



          Recall that by Binomial Expansion,



          $$(a+b)^n = sum_{r = 0}^{n} {n choose r}a^{n-r}b^r$$



          So you get



          $$lim_{h rightarrow 0} {color{blue}{{99 choose 0}x^{99}}+{99 choose 1}x^{98}h+{99 choose 2}x^{97}h^2+…+{99 choose n}h^{99}color{red}{-x^{99}}over h}$$



          Cancel out the first and last terms. From here, notice if anything can be factored. Also, as another hint (for later simplifications), $${n choose r} = frac{n!}{r!{(n-r)!}}$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 26 '18 at 22:10

























          answered Nov 26 '18 at 22:05









          KM101KM101

          5,9281523




          5,9281523























              1












              $begingroup$

              The trick is to realise that $h to 0$ means $h$ is arbitrarily small so we can assume $h < 1$ and so $h > h^2 > h^3$ and that the "higher values of $h$" become "negligible"



              So you are thinking way too hard. Just use the binomial theorem to get $(x + h)^{99} = x^{99} + 99hx^{98} + sum_{k=2}^{99} {99 choose k} h^kx^{99-k}$ and hope that must of those higher values will become negligible.



              And indeed:



              $require{cancel}limlimits_{hto 0}frac {(x + h)^{99}-x^{99}}{h }=$



              $limlimits_{hto 0}frac {x^{99} + 99hx^{98} + (sum_{k=2}^{99} {99 choose k} h^kx^{99-k})-x^{99}}{h}=$



              $limlimits_{hto 0}frac {color{green}{cancel{x^{99}}} + 99color{red}{cancel h}x^{98} + (sum_{k=2}^{99} {99 choose k} h^{color{red}{cancel {k}}k-1}x^{99-k}) - color{green}{cancel{x^{99}}}}{color{red}{cancel h}}=$



              $limlimits_{hto 0}(99x^{98} + sum_{k=2}^{99} {99 choose k}h^{k-1}x^{99-k})=$



              $99x^{98} + sum_{k=2}^{99}{99 choose k}(limlimits_{hto 0} h^{k-1})x^{99-k}=$



              $99x^{98} + sum_{k=2}^{99}{99 choose k}(0)x^{99-k}=$



              $99x^{98} + sum_{k=2}^{99}0=$



              $99x^{98}$



              ======



              Doug M had an intriguing hint in the comments:



              Considering $a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b+ .... + ab^{n-2}+ b^{99})$ we get:



              $(x + h)^{99} - x^{99}=$



              $[(x+h) - x][(x+h)^{98} + (x+h)^{97}x + ... + (x+h)x^{97} + x^{98}]=$



              $h[(x+h)^{98} + (x+h)^{97}x + ... + (x+h)x^{97} + x^{98}]$



              And



              $[(x+h)^{98} + (x+h)^{97}x + ... + (x+h)x^{97} + x^{98}]=$



              $(x^{98} + text{a bunch of stuff with h as a factor})+(x^{97}cdot x + x(text{a bunch of stuff with h as a factor}))+... (xcdot x^{97} + x^{97}(text{a bunch of stuff with h as a factor})) + x^{98})=$



              $99x^{98} + text { a REALLY big bunch of stuff with h as a factor}$



              So $frac {(x+h)^{98} - x^{99}}h=$



              $frac {h(99x^{98}+ text { a REALLY big bunch of stuff with h as a factor}}h = $



              $99x^{98} + text { a REALLY big bunch of stuff with h as a factor}$



              ... and "$ text { a REALLY big bunch of stuff with h as a factor}$" goes to $0$ as .... it's a REALLY big bunch of stuff with $h$ as a factor.



              Okay, .... it was an intriguing idea and I'm glad I did it but.... I don't think it's a very practical way to do it.






              share|cite|improve this answer











              $endgroup$









              • 1




                $begingroup$
                Hoping that you don't mind, I shall adopt and reuse the REALLY big bunch of stuff
                $endgroup$
                – Claude Leibovici
                Nov 27 '18 at 6:01
















              1












              $begingroup$

              The trick is to realise that $h to 0$ means $h$ is arbitrarily small so we can assume $h < 1$ and so $h > h^2 > h^3$ and that the "higher values of $h$" become "negligible"



              So you are thinking way too hard. Just use the binomial theorem to get $(x + h)^{99} = x^{99} + 99hx^{98} + sum_{k=2}^{99} {99 choose k} h^kx^{99-k}$ and hope that must of those higher values will become negligible.



              And indeed:



              $require{cancel}limlimits_{hto 0}frac {(x + h)^{99}-x^{99}}{h }=$



              $limlimits_{hto 0}frac {x^{99} + 99hx^{98} + (sum_{k=2}^{99} {99 choose k} h^kx^{99-k})-x^{99}}{h}=$



              $limlimits_{hto 0}frac {color{green}{cancel{x^{99}}} + 99color{red}{cancel h}x^{98} + (sum_{k=2}^{99} {99 choose k} h^{color{red}{cancel {k}}k-1}x^{99-k}) - color{green}{cancel{x^{99}}}}{color{red}{cancel h}}=$



              $limlimits_{hto 0}(99x^{98} + sum_{k=2}^{99} {99 choose k}h^{k-1}x^{99-k})=$



              $99x^{98} + sum_{k=2}^{99}{99 choose k}(limlimits_{hto 0} h^{k-1})x^{99-k}=$



              $99x^{98} + sum_{k=2}^{99}{99 choose k}(0)x^{99-k}=$



              $99x^{98} + sum_{k=2}^{99}0=$



              $99x^{98}$



              ======



              Doug M had an intriguing hint in the comments:



              Considering $a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b+ .... + ab^{n-2}+ b^{99})$ we get:



              $(x + h)^{99} - x^{99}=$



              $[(x+h) - x][(x+h)^{98} + (x+h)^{97}x + ... + (x+h)x^{97} + x^{98}]=$



              $h[(x+h)^{98} + (x+h)^{97}x + ... + (x+h)x^{97} + x^{98}]$



              And



              $[(x+h)^{98} + (x+h)^{97}x + ... + (x+h)x^{97} + x^{98}]=$



              $(x^{98} + text{a bunch of stuff with h as a factor})+(x^{97}cdot x + x(text{a bunch of stuff with h as a factor}))+... (xcdot x^{97} + x^{97}(text{a bunch of stuff with h as a factor})) + x^{98})=$



              $99x^{98} + text { a REALLY big bunch of stuff with h as a factor}$



              So $frac {(x+h)^{98} - x^{99}}h=$



              $frac {h(99x^{98}+ text { a REALLY big bunch of stuff with h as a factor}}h = $



              $99x^{98} + text { a REALLY big bunch of stuff with h as a factor}$



              ... and "$ text { a REALLY big bunch of stuff with h as a factor}$" goes to $0$ as .... it's a REALLY big bunch of stuff with $h$ as a factor.



              Okay, .... it was an intriguing idea and I'm glad I did it but.... I don't think it's a very practical way to do it.






              share|cite|improve this answer











              $endgroup$









              • 1




                $begingroup$
                Hoping that you don't mind, I shall adopt and reuse the REALLY big bunch of stuff
                $endgroup$
                – Claude Leibovici
                Nov 27 '18 at 6:01














              1












              1








              1





              $begingroup$

              The trick is to realise that $h to 0$ means $h$ is arbitrarily small so we can assume $h < 1$ and so $h > h^2 > h^3$ and that the "higher values of $h$" become "negligible"



              So you are thinking way too hard. Just use the binomial theorem to get $(x + h)^{99} = x^{99} + 99hx^{98} + sum_{k=2}^{99} {99 choose k} h^kx^{99-k}$ and hope that must of those higher values will become negligible.



              And indeed:



              $require{cancel}limlimits_{hto 0}frac {(x + h)^{99}-x^{99}}{h }=$



              $limlimits_{hto 0}frac {x^{99} + 99hx^{98} + (sum_{k=2}^{99} {99 choose k} h^kx^{99-k})-x^{99}}{h}=$



              $limlimits_{hto 0}frac {color{green}{cancel{x^{99}}} + 99color{red}{cancel h}x^{98} + (sum_{k=2}^{99} {99 choose k} h^{color{red}{cancel {k}}k-1}x^{99-k}) - color{green}{cancel{x^{99}}}}{color{red}{cancel h}}=$



              $limlimits_{hto 0}(99x^{98} + sum_{k=2}^{99} {99 choose k}h^{k-1}x^{99-k})=$



              $99x^{98} + sum_{k=2}^{99}{99 choose k}(limlimits_{hto 0} h^{k-1})x^{99-k}=$



              $99x^{98} + sum_{k=2}^{99}{99 choose k}(0)x^{99-k}=$



              $99x^{98} + sum_{k=2}^{99}0=$



              $99x^{98}$



              ======



              Doug M had an intriguing hint in the comments:



              Considering $a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b+ .... + ab^{n-2}+ b^{99})$ we get:



              $(x + h)^{99} - x^{99}=$



              $[(x+h) - x][(x+h)^{98} + (x+h)^{97}x + ... + (x+h)x^{97} + x^{98}]=$



              $h[(x+h)^{98} + (x+h)^{97}x + ... + (x+h)x^{97} + x^{98}]$



              And



              $[(x+h)^{98} + (x+h)^{97}x + ... + (x+h)x^{97} + x^{98}]=$



              $(x^{98} + text{a bunch of stuff with h as a factor})+(x^{97}cdot x + x(text{a bunch of stuff with h as a factor}))+... (xcdot x^{97} + x^{97}(text{a bunch of stuff with h as a factor})) + x^{98})=$



              $99x^{98} + text { a REALLY big bunch of stuff with h as a factor}$



              So $frac {(x+h)^{98} - x^{99}}h=$



              $frac {h(99x^{98}+ text { a REALLY big bunch of stuff with h as a factor}}h = $



              $99x^{98} + text { a REALLY big bunch of stuff with h as a factor}$



              ... and "$ text { a REALLY big bunch of stuff with h as a factor}$" goes to $0$ as .... it's a REALLY big bunch of stuff with $h$ as a factor.



              Okay, .... it was an intriguing idea and I'm glad I did it but.... I don't think it's a very practical way to do it.






              share|cite|improve this answer











              $endgroup$



              The trick is to realise that $h to 0$ means $h$ is arbitrarily small so we can assume $h < 1$ and so $h > h^2 > h^3$ and that the "higher values of $h$" become "negligible"



              So you are thinking way too hard. Just use the binomial theorem to get $(x + h)^{99} = x^{99} + 99hx^{98} + sum_{k=2}^{99} {99 choose k} h^kx^{99-k}$ and hope that must of those higher values will become negligible.



              And indeed:



              $require{cancel}limlimits_{hto 0}frac {(x + h)^{99}-x^{99}}{h }=$



              $limlimits_{hto 0}frac {x^{99} + 99hx^{98} + (sum_{k=2}^{99} {99 choose k} h^kx^{99-k})-x^{99}}{h}=$



              $limlimits_{hto 0}frac {color{green}{cancel{x^{99}}} + 99color{red}{cancel h}x^{98} + (sum_{k=2}^{99} {99 choose k} h^{color{red}{cancel {k}}k-1}x^{99-k}) - color{green}{cancel{x^{99}}}}{color{red}{cancel h}}=$



              $limlimits_{hto 0}(99x^{98} + sum_{k=2}^{99} {99 choose k}h^{k-1}x^{99-k})=$



              $99x^{98} + sum_{k=2}^{99}{99 choose k}(limlimits_{hto 0} h^{k-1})x^{99-k}=$



              $99x^{98} + sum_{k=2}^{99}{99 choose k}(0)x^{99-k}=$



              $99x^{98} + sum_{k=2}^{99}0=$



              $99x^{98}$



              ======



              Doug M had an intriguing hint in the comments:



              Considering $a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b+ .... + ab^{n-2}+ b^{99})$ we get:



              $(x + h)^{99} - x^{99}=$



              $[(x+h) - x][(x+h)^{98} + (x+h)^{97}x + ... + (x+h)x^{97} + x^{98}]=$



              $h[(x+h)^{98} + (x+h)^{97}x + ... + (x+h)x^{97} + x^{98}]$



              And



              $[(x+h)^{98} + (x+h)^{97}x + ... + (x+h)x^{97} + x^{98}]=$



              $(x^{98} + text{a bunch of stuff with h as a factor})+(x^{97}cdot x + x(text{a bunch of stuff with h as a factor}))+... (xcdot x^{97} + x^{97}(text{a bunch of stuff with h as a factor})) + x^{98})=$



              $99x^{98} + text { a REALLY big bunch of stuff with h as a factor}$



              So $frac {(x+h)^{98} - x^{99}}h=$



              $frac {h(99x^{98}+ text { a REALLY big bunch of stuff with h as a factor}}h = $



              $99x^{98} + text { a REALLY big bunch of stuff with h as a factor}$



              ... and "$ text { a REALLY big bunch of stuff with h as a factor}$" goes to $0$ as .... it's a REALLY big bunch of stuff with $h$ as a factor.



              Okay, .... it was an intriguing idea and I'm glad I did it but.... I don't think it's a very practical way to do it.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Nov 26 '18 at 23:12

























              answered Nov 26 '18 at 22:53









              fleabloodfleablood

              69.3k22685




              69.3k22685








              • 1




                $begingroup$
                Hoping that you don't mind, I shall adopt and reuse the REALLY big bunch of stuff
                $endgroup$
                – Claude Leibovici
                Nov 27 '18 at 6:01














              • 1




                $begingroup$
                Hoping that you don't mind, I shall adopt and reuse the REALLY big bunch of stuff
                $endgroup$
                – Claude Leibovici
                Nov 27 '18 at 6:01








              1




              1




              $begingroup$
              Hoping that you don't mind, I shall adopt and reuse the REALLY big bunch of stuff
              $endgroup$
              – Claude Leibovici
              Nov 27 '18 at 6:01




              $begingroup$
              Hoping that you don't mind, I shall adopt and reuse the REALLY big bunch of stuff
              $endgroup$
              – Claude Leibovici
              Nov 27 '18 at 6:01











              0












              $begingroup$

              Because both $limlimits_{hto 0}((x-h)^{99}-x^{99})=0$ and $limlimits_{hto 0}h=0$ you may also use L'Hospital's rule:



              $displaystylelimlimits_{hto 0}frac{(x-h)^{99}-x^{99}}{h}=displaystylelimlimits_{hto 0}frac{frac{d}{dh}((x-h)^{99}-x^{99})}{frac{d}{dh}h}=displaystylelimlimits_{hto 0}frac{99(x-h)^{98}-0}{1}=99x^{98}$






              share|cite|improve this answer









              $endgroup$









              • 1




                $begingroup$
                L'Hospital is far too complicated. If you know derivatives, you can use the definition in this case.
                $endgroup$
                – gammatester
                Nov 26 '18 at 22:05
















              0












              $begingroup$

              Because both $limlimits_{hto 0}((x-h)^{99}-x^{99})=0$ and $limlimits_{hto 0}h=0$ you may also use L'Hospital's rule:



              $displaystylelimlimits_{hto 0}frac{(x-h)^{99}-x^{99}}{h}=displaystylelimlimits_{hto 0}frac{frac{d}{dh}((x-h)^{99}-x^{99})}{frac{d}{dh}h}=displaystylelimlimits_{hto 0}frac{99(x-h)^{98}-0}{1}=99x^{98}$






              share|cite|improve this answer









              $endgroup$









              • 1




                $begingroup$
                L'Hospital is far too complicated. If you know derivatives, you can use the definition in this case.
                $endgroup$
                – gammatester
                Nov 26 '18 at 22:05














              0












              0








              0





              $begingroup$

              Because both $limlimits_{hto 0}((x-h)^{99}-x^{99})=0$ and $limlimits_{hto 0}h=0$ you may also use L'Hospital's rule:



              $displaystylelimlimits_{hto 0}frac{(x-h)^{99}-x^{99}}{h}=displaystylelimlimits_{hto 0}frac{frac{d}{dh}((x-h)^{99}-x^{99})}{frac{d}{dh}h}=displaystylelimlimits_{hto 0}frac{99(x-h)^{98}-0}{1}=99x^{98}$






              share|cite|improve this answer









              $endgroup$



              Because both $limlimits_{hto 0}((x-h)^{99}-x^{99})=0$ and $limlimits_{hto 0}h=0$ you may also use L'Hospital's rule:



              $displaystylelimlimits_{hto 0}frac{(x-h)^{99}-x^{99}}{h}=displaystylelimlimits_{hto 0}frac{frac{d}{dh}((x-h)^{99}-x^{99})}{frac{d}{dh}h}=displaystylelimlimits_{hto 0}frac{99(x-h)^{98}-0}{1}=99x^{98}$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 26 '18 at 22:03









              Martin RosenauMartin Rosenau

              1,156139




              1,156139








              • 1




                $begingroup$
                L'Hospital is far too complicated. If you know derivatives, you can use the definition in this case.
                $endgroup$
                – gammatester
                Nov 26 '18 at 22:05














              • 1




                $begingroup$
                L'Hospital is far too complicated. If you know derivatives, you can use the definition in this case.
                $endgroup$
                – gammatester
                Nov 26 '18 at 22:05








              1




              1




              $begingroup$
              L'Hospital is far too complicated. If you know derivatives, you can use the definition in this case.
              $endgroup$
              – gammatester
              Nov 26 '18 at 22:05




              $begingroup$
              L'Hospital is far too complicated. If you know derivatives, you can use the definition in this case.
              $endgroup$
              – gammatester
              Nov 26 '18 at 22:05











              0












              $begingroup$

              This limit is the derivative of $f(x)=x^{99}$. To get a generalization you can replace $99$ with $n$ where $n in mathbb{R}$. The limit thus translates to $displaystyle lim_{h to 0} frac{(x+h)^{n}-x^n}{h}$.



              Using Binomial Theorem, which states that $(a+b)^n=displaystyle sum_{k=0}^n {n choose k} a^{n-k} b^k$. Substituting $begin{pmatrix}a \ bend{pmatrix}=begin{pmatrix}x \ hend{pmatrix}$. $$(x+h)^n=x^n+{n choose 1}x^{n-1} h + {n choose 2} x^{n-2} h^2+ cdots +{n choose n-1}xh^{n-1}+h^n$$ Putting back this result in the expression for the limit. We get the following limit $$displaystyle lim_{h to 0} frac{(x+h)^n-x^n}{h}=displaystyle lim_{h to 0} Biggl(frac{1}{h}{n choose 1}x^{n-1}h+frac{1}{h}biggl(h^2biggl({n choose 2}x^{n-2}+{n choose 3}x^{n-3}h+cdots+h^{n-2}biggr)biggr)Biggr)$$ $$=displaystyle lim_{h to 0} Biggl(biggl(nx^{n-1}biggr)+hbiggl({n choose 2}x^{n-2}+{n choose 3}x^{n-3}h+cdots+h^{n-2}biggr)Biggr)$$ $$=displaystyle lim_{h to 0}nx^{n-1}+underbrace{displaystyle lim_{h to 0}hbiggl({n choose 2}x^{n-2}+{n choose 3}x^{n-3}h+cdots+h^{n-2}biggr)}_{= 0}$$
              $$implies displaystyle lim_{h to 0}frac{(x+h)^n-x^n}{h}= nx^{n-1}$$
              Therefore, For $n=99$, the given limit evaluates to $99x^{98}$. Cheers






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                This limit is the derivative of $f(x)=x^{99}$. To get a generalization you can replace $99$ with $n$ where $n in mathbb{R}$. The limit thus translates to $displaystyle lim_{h to 0} frac{(x+h)^{n}-x^n}{h}$.



                Using Binomial Theorem, which states that $(a+b)^n=displaystyle sum_{k=0}^n {n choose k} a^{n-k} b^k$. Substituting $begin{pmatrix}a \ bend{pmatrix}=begin{pmatrix}x \ hend{pmatrix}$. $$(x+h)^n=x^n+{n choose 1}x^{n-1} h + {n choose 2} x^{n-2} h^2+ cdots +{n choose n-1}xh^{n-1}+h^n$$ Putting back this result in the expression for the limit. We get the following limit $$displaystyle lim_{h to 0} frac{(x+h)^n-x^n}{h}=displaystyle lim_{h to 0} Biggl(frac{1}{h}{n choose 1}x^{n-1}h+frac{1}{h}biggl(h^2biggl({n choose 2}x^{n-2}+{n choose 3}x^{n-3}h+cdots+h^{n-2}biggr)biggr)Biggr)$$ $$=displaystyle lim_{h to 0} Biggl(biggl(nx^{n-1}biggr)+hbiggl({n choose 2}x^{n-2}+{n choose 3}x^{n-3}h+cdots+h^{n-2}biggr)Biggr)$$ $$=displaystyle lim_{h to 0}nx^{n-1}+underbrace{displaystyle lim_{h to 0}hbiggl({n choose 2}x^{n-2}+{n choose 3}x^{n-3}h+cdots+h^{n-2}biggr)}_{= 0}$$
                $$implies displaystyle lim_{h to 0}frac{(x+h)^n-x^n}{h}= nx^{n-1}$$
                Therefore, For $n=99$, the given limit evaluates to $99x^{98}$. Cheers






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  This limit is the derivative of $f(x)=x^{99}$. To get a generalization you can replace $99$ with $n$ where $n in mathbb{R}$. The limit thus translates to $displaystyle lim_{h to 0} frac{(x+h)^{n}-x^n}{h}$.



                  Using Binomial Theorem, which states that $(a+b)^n=displaystyle sum_{k=0}^n {n choose k} a^{n-k} b^k$. Substituting $begin{pmatrix}a \ bend{pmatrix}=begin{pmatrix}x \ hend{pmatrix}$. $$(x+h)^n=x^n+{n choose 1}x^{n-1} h + {n choose 2} x^{n-2} h^2+ cdots +{n choose n-1}xh^{n-1}+h^n$$ Putting back this result in the expression for the limit. We get the following limit $$displaystyle lim_{h to 0} frac{(x+h)^n-x^n}{h}=displaystyle lim_{h to 0} Biggl(frac{1}{h}{n choose 1}x^{n-1}h+frac{1}{h}biggl(h^2biggl({n choose 2}x^{n-2}+{n choose 3}x^{n-3}h+cdots+h^{n-2}biggr)biggr)Biggr)$$ $$=displaystyle lim_{h to 0} Biggl(biggl(nx^{n-1}biggr)+hbiggl({n choose 2}x^{n-2}+{n choose 3}x^{n-3}h+cdots+h^{n-2}biggr)Biggr)$$ $$=displaystyle lim_{h to 0}nx^{n-1}+underbrace{displaystyle lim_{h to 0}hbiggl({n choose 2}x^{n-2}+{n choose 3}x^{n-3}h+cdots+h^{n-2}biggr)}_{= 0}$$
                  $$implies displaystyle lim_{h to 0}frac{(x+h)^n-x^n}{h}= nx^{n-1}$$
                  Therefore, For $n=99$, the given limit evaluates to $99x^{98}$. Cheers






                  share|cite|improve this answer









                  $endgroup$



                  This limit is the derivative of $f(x)=x^{99}$. To get a generalization you can replace $99$ with $n$ where $n in mathbb{R}$. The limit thus translates to $displaystyle lim_{h to 0} frac{(x+h)^{n}-x^n}{h}$.



                  Using Binomial Theorem, which states that $(a+b)^n=displaystyle sum_{k=0}^n {n choose k} a^{n-k} b^k$. Substituting $begin{pmatrix}a \ bend{pmatrix}=begin{pmatrix}x \ hend{pmatrix}$. $$(x+h)^n=x^n+{n choose 1}x^{n-1} h + {n choose 2} x^{n-2} h^2+ cdots +{n choose n-1}xh^{n-1}+h^n$$ Putting back this result in the expression for the limit. We get the following limit $$displaystyle lim_{h to 0} frac{(x+h)^n-x^n}{h}=displaystyle lim_{h to 0} Biggl(frac{1}{h}{n choose 1}x^{n-1}h+frac{1}{h}biggl(h^2biggl({n choose 2}x^{n-2}+{n choose 3}x^{n-3}h+cdots+h^{n-2}biggr)biggr)Biggr)$$ $$=displaystyle lim_{h to 0} Biggl(biggl(nx^{n-1}biggr)+hbiggl({n choose 2}x^{n-2}+{n choose 3}x^{n-3}h+cdots+h^{n-2}biggr)Biggr)$$ $$=displaystyle lim_{h to 0}nx^{n-1}+underbrace{displaystyle lim_{h to 0}hbiggl({n choose 2}x^{n-2}+{n choose 3}x^{n-3}h+cdots+h^{n-2}biggr)}_{= 0}$$
                  $$implies displaystyle lim_{h to 0}frac{(x+h)^n-x^n}{h}= nx^{n-1}$$
                  Therefore, For $n=99$, the given limit evaluates to $99x^{98}$. Cheers







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                  answered Nov 28 '18 at 8:18









                  Paras KhoslaParas Khosla

                  8610




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