Evaluate $lim_{h rightarrow 0} {(x+h)^{99}-x^{99}over h}$
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$$lim_{h rightarrow 0} {(x+h)^{99}-x^{99}over h}$$
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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|
show 2 more comments
$begingroup$
$$lim_{h rightarrow 0} {(x+h)^{99}-x^{99}over h}$$
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?
algebra-precalculus limits functions
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Do you mean $h to 0$?
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– MisterRiemann
Nov 26 '18 at 21:52
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yes, thank you @MisterRiemann
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– didgocks
Nov 26 '18 at 21:53
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Hint: Limit definition of the derivative of $f(x)=x^{99}$.
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– Anurag A
Nov 26 '18 at 21:53
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Use the binomial theorem.
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– gammatester
Nov 26 '18 at 21:54
1
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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 $a = x$
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– Doug M
Nov 26 '18 at 21:57
|
show 2 more comments
$begingroup$
$$lim_{h rightarrow 0} {(x+h)^{99}-x^{99}over h}$$
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?
algebra-precalculus limits functions
$endgroup$
$$lim_{h rightarrow 0} {(x+h)^{99}-x^{99}over h}$$
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
algebra-precalculus limits functions
edited Nov 27 '18 at 4:44
Paramanand Singh
49.6k556163
49.6k556163
asked Nov 26 '18 at 21:51
didgocksdidgocks
6801823
6801823
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Do you mean $h to 0$?
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– MisterRiemann
Nov 26 '18 at 21:52
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yes, thank you @MisterRiemann
$endgroup$
– didgocks
Nov 26 '18 at 21:53
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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 $a = x$
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– Doug M
Nov 26 '18 at 21:57
|
show 2 more comments
$begingroup$
Do you mean $h to 0$?
$endgroup$
– MisterRiemann
Nov 26 '18 at 21:52
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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
|
show 2 more comments
5 Answers
5
active
oldest
votes
$begingroup$
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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1
$begingroup$
You made a typing mistake: It should be $f(x+h)-f(x)$, not $f(x+h)-f(h)$.
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– Martin Rosenau
Nov 26 '18 at 21:58
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@MartinRosenau Thanks!
$endgroup$
– MisterRiemann
Nov 26 '18 at 21:58
add a comment |
$begingroup$
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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add a comment |
$begingroup$
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.
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1
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Hoping that you don't mind, I shall adopt and reuse the REALLY big bunch of stuff
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– Claude Leibovici
Nov 27 '18 at 6:01
add a comment |
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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:
$displaystylelimlimits_{hto 0}frac{(x-h)^{99}-x^{99}}{h}=displaystylelimlimits_{hto 0}frac{frac{d}{dh}((x-h)^{99}-x^{99})}{frac{d}{dh}h}=displaystylelimlimits_{hto 0}frac{99(x-h)^{98}-0}{1}=99x^{98}$
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1
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L'Hospital is far too complicated. If you know derivatives, you can use the definition in this case.
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– gammatester
Nov 26 '18 at 22:05
add a comment |
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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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add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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}
$endgroup$
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
add a comment |
$begingroup$
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}
$endgroup$
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
add a comment |
$begingroup$
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}
$endgroup$
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}
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
add a comment |
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
add a comment |
$begingroup$
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), $${n choose r} = frac{n!}{r!{(n-r)!}}$$
$endgroup$
add a comment |
$begingroup$
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), $${n choose r} = frac{n!}{r!{(n-r)!}}$$
$endgroup$
add a comment |
$begingroup$
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), $${n choose r} = frac{n!}{r!{(n-r)!}}$$
$endgroup$
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), $${n choose r} = frac{n!}{r!{(n-r)!}}$$
edited Nov 26 '18 at 22:10
answered Nov 26 '18 at 22:05
KM101KM101
5,9281523
5,9281523
add a comment |
add a comment |
$begingroup$
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.
$endgroup$
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
add a comment |
$begingroup$
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.
$endgroup$
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
add a comment |
$begingroup$
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.
$endgroup$
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.
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
add a comment |
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
add a comment |
$begingroup$
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:
$displaystylelimlimits_{hto 0}frac{(x-h)^{99}-x^{99}}{h}=displaystylelimlimits_{hto 0}frac{frac{d}{dh}((x-h)^{99}-x^{99})}{frac{d}{dh}h}=displaystylelimlimits_{hto 0}frac{99(x-h)^{98}-0}{1}=99x^{98}$
$endgroup$
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
add a comment |
$begingroup$
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:
$displaystylelimlimits_{hto 0}frac{(x-h)^{99}-x^{99}}{h}=displaystylelimlimits_{hto 0}frac{frac{d}{dh}((x-h)^{99}-x^{99})}{frac{d}{dh}h}=displaystylelimlimits_{hto 0}frac{99(x-h)^{98}-0}{1}=99x^{98}$
$endgroup$
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
add a comment |
$begingroup$
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:
$displaystylelimlimits_{hto 0}frac{(x-h)^{99}-x^{99}}{h}=displaystylelimlimits_{hto 0}frac{frac{d}{dh}((x-h)^{99}-x^{99})}{frac{d}{dh}h}=displaystylelimlimits_{hto 0}frac{99(x-h)^{98}-0}{1}=99x^{98}$
$endgroup$
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:
$displaystylelimlimits_{hto 0}frac{(x-h)^{99}-x^{99}}{h}=displaystylelimlimits_{hto 0}frac{frac{d}{dh}((x-h)^{99}-x^{99})}{frac{d}{dh}h}=displaystylelimlimits_{hto 0}frac{99(x-h)^{98}-0}{1}=99x^{98}$
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
add a comment |
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
add a comment |
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
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
answered Nov 28 '18 at 8:18
Paras KhoslaParas Khosla
8610
8610
add a comment |
add a comment |
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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