Proof that $exp(x+y) = exp(x)*exp(y)$ using limit definition of $exp(x)$












4












$begingroup$


I want to prove: $exp(x+y) = exp(x)cdot exp(y)$ using the definition: $exp(x) = lim_{ntoinfty} (1+frac{x}{n})^n$



I am having trouble completing the proof, but here is my idea so far: $$lim_{ntoinfty} left(1+frac{x+y}{n}right)^n = lim_{ntoinfty} left(1+frac{x}{n}right)^n cdot lim_{ntoinfty} left(1+frac{y}{n}right)^n =
lim_{ntoinfty} left(left(1+frac{x}{n}right)^n cdot left(1+frac{y}{n}right)^n right) $$



Now I rearrange the last expression: $$lim_{ntoinfty} left(1+frac{x+y+frac{xy}{n}}{n}right)^n $$



From here my idea is to somehow show that this limit is equal to $$lim_{ntoinfty} left(1+frac{x+y}{n}right)^n = exp(x+y)$$ using the Squeeze Theorem and perhaps Bernoulli's Inequality, but I am at a loss as to how exactly to do it. I'd appreciate your help.






real-analysis limits exponential-function







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$endgroup$








  • 9




    $begingroup$
    These should be the limit as $nrightarrowinfty$?
    $endgroup$
    – gd1035
    Nov 27 '18 at 18:59










  • $begingroup$
    Thank you, yes. Fixed.
    $endgroup$
    – TomRatTUM
    Nov 27 '18 at 19:16










  • $begingroup$
    See this answer math.stackexchange.com/a/3000717/72031
    $endgroup$
    – Paramanand Singh
    Nov 28 '18 at 1:50
















4












$begingroup$


I want to prove: $exp(x+y) = exp(x)cdot exp(y)$ using the definition: $exp(x) = lim_{ntoinfty} (1+frac{x}{n})^n$



I am having trouble completing the proof, but here is my idea so far: $$lim_{ntoinfty} left(1+frac{x+y}{n}right)^n = lim_{ntoinfty} left(1+frac{x}{n}right)^n cdot lim_{ntoinfty} left(1+frac{y}{n}right)^n =
lim_{ntoinfty} left(left(1+frac{x}{n}right)^n cdot left(1+frac{y}{n}right)^n right) $$



Now I rearrange the last expression: $$lim_{ntoinfty} left(1+frac{x+y+frac{xy}{n}}{n}right)^n $$



From here my idea is to somehow show that this limit is equal to $$lim_{ntoinfty} left(1+frac{x+y}{n}right)^n = exp(x+y)$$ using the Squeeze Theorem and perhaps Bernoulli's Inequality, but I am at a loss as to how exactly to do it. I'd appreciate your help.






real-analysis limits exponential-function







share|cite|improve this question











$endgroup$








  • 9




    $begingroup$
    These should be the limit as $nrightarrowinfty$?
    $endgroup$
    – gd1035
    Nov 27 '18 at 18:59










  • $begingroup$
    Thank you, yes. Fixed.
    $endgroup$
    – TomRatTUM
    Nov 27 '18 at 19:16










  • $begingroup$
    See this answer math.stackexchange.com/a/3000717/72031
    $endgroup$
    – Paramanand Singh
    Nov 28 '18 at 1:50














4












4








4





$begingroup$


I want to prove: $exp(x+y) = exp(x)cdot exp(y)$ using the definition: $exp(x) = lim_{ntoinfty} (1+frac{x}{n})^n$



I am having trouble completing the proof, but here is my idea so far: $$lim_{ntoinfty} left(1+frac{x+y}{n}right)^n = lim_{ntoinfty} left(1+frac{x}{n}right)^n cdot lim_{ntoinfty} left(1+frac{y}{n}right)^n =
lim_{ntoinfty} left(left(1+frac{x}{n}right)^n cdot left(1+frac{y}{n}right)^n right) $$



Now I rearrange the last expression: $$lim_{ntoinfty} left(1+frac{x+y+frac{xy}{n}}{n}right)^n $$



From here my idea is to somehow show that this limit is equal to $$lim_{ntoinfty} left(1+frac{x+y}{n}right)^n = exp(x+y)$$ using the Squeeze Theorem and perhaps Bernoulli's Inequality, but I am at a loss as to how exactly to do it. I'd appreciate your help.






real-analysis limits exponential-function







share|cite|improve this question











$endgroup$




I want to prove: $exp(x+y) = exp(x)cdot exp(y)$ using the definition: $exp(x) = lim_{ntoinfty} (1+frac{x}{n})^n$



I am having trouble completing the proof, but here is my idea so far: $$lim_{ntoinfty} left(1+frac{x+y}{n}right)^n = lim_{ntoinfty} left(1+frac{x}{n}right)^n cdot lim_{ntoinfty} left(1+frac{y}{n}right)^n =
lim_{ntoinfty} left(left(1+frac{x}{n}right)^n cdot left(1+frac{y}{n}right)^n right) $$



Now I rearrange the last expression: $$lim_{ntoinfty} left(1+frac{x+y+frac{xy}{n}}{n}right)^n $$



From here my idea is to somehow show that this limit is equal to $$lim_{ntoinfty} left(1+frac{x+y}{n}right)^n = exp(x+y)$$ using the Squeeze Theorem and perhaps Bernoulli's Inequality, but I am at a loss as to how exactly to do it. I'd appreciate your help.






real-analysis limits exponential-function




real-analysis limits exponential-function






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 '18 at 19:15







TomRatTUM

















asked Nov 27 '18 at 18:57









TomRatTUMTomRatTUM

234




234








  • 9




    $begingroup$
    These should be the limit as $nrightarrowinfty$?
    $endgroup$
    – gd1035
    Nov 27 '18 at 18:59










  • $begingroup$
    Thank you, yes. Fixed.
    $endgroup$
    – TomRatTUM
    Nov 27 '18 at 19:16










  • $begingroup$
    See this answer math.stackexchange.com/a/3000717/72031
    $endgroup$
    – Paramanand Singh
    Nov 28 '18 at 1:50














  • 9




    $begingroup$
    These should be the limit as $nrightarrowinfty$?
    $endgroup$
    – gd1035
    Nov 27 '18 at 18:59










  • $begingroup$
    Thank you, yes. Fixed.
    $endgroup$
    – TomRatTUM
    Nov 27 '18 at 19:16










  • $begingroup$
    See this answer math.stackexchange.com/a/3000717/72031
    $endgroup$
    – Paramanand Singh
    Nov 28 '18 at 1:50








9




9




$begingroup$
These should be the limit as $nrightarrowinfty$?
$endgroup$
– gd1035
Nov 27 '18 at 18:59




$begingroup$
These should be the limit as $nrightarrowinfty$?
$endgroup$
– gd1035
Nov 27 '18 at 18:59












$begingroup$
Thank you, yes. Fixed.
$endgroup$
– TomRatTUM
Nov 27 '18 at 19:16




$begingroup$
Thank you, yes. Fixed.
$endgroup$
– TomRatTUM
Nov 27 '18 at 19:16












$begingroup$
See this answer math.stackexchange.com/a/3000717/72031
$endgroup$
– Paramanand Singh
Nov 28 '18 at 1:50




$begingroup$
See this answer math.stackexchange.com/a/3000717/72031
$endgroup$
– Paramanand Singh
Nov 28 '18 at 1:50










2 Answers
2






active

oldest

votes


















1












$begingroup$

$$
\dfrac{e^{x+y}}{e^x}=lim_{nto+infty}{Big(dfrac{1+frac {x+y} n}{1+frac x n}Big)^n}=
\lim_{nto+infty}Big(frac{x+y+n}{x+n}Big)^n=lim_{nto+infty}Big(1+frac y{x+n}Big)^n=
\lim_{nto+infty}frac{Big(1+frac y {x+n}Big)^{n+x}}{Big(1+frac y {x+n}Big)^x}=e^y
$$
Because $lim_{nto+infty}{Big(1+frac y {x+n}Big)^{n+x}}=e^y$ and $lim_{nto+infty}{Big(1+frac y {x+n}Big)^x}=1$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    $x+n$ is not necessarily a positive integer.
    $endgroup$
    – Paramanand Singh
    Nov 28 '18 at 1:52



















2












$begingroup$

It suffices to show that
$$
lim_{ntoinfty}frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=1 tag{1}
$$
But
$$
frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=
left(1+frac{xy}{n^2(1+frac{x+y}{n})}right)^n
$$
and as, for suitably large $n$, say $nge n_0$, we have that
$$
frac{1}{2}<1+frac{x+y}{n}<2,
$$
then
$$
left(1+frac{xy}{2n^2}right)^n<frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=
left(1+frac{xy}{n^2(1+frac{x+y}{n})}right)^n<left(1+frac{2xy}{n^2}right)^n
$$
Now, for all $zinmathbb R$, we have
$$
left(1+frac{z}{n^2}right)^n=left(left(1+frac{z}{n^2}right)^{n^2}right)^{1/n}to 1,
$$
since $left(1+frac{z}{n^2}right)^{n^2}to e^z$.



Thus
$$
lim_{ntoinfty}left(1+frac{xy}{2n^2}right)^n=lim_{ntoinfty}left(1+frac{2xy}{n^2}right)^n=1,
$$
and hence $(1)$ holds.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    $$
    \dfrac{e^{x+y}}{e^x}=lim_{nto+infty}{Big(dfrac{1+frac {x+y} n}{1+frac x n}Big)^n}=
    \lim_{nto+infty}Big(frac{x+y+n}{x+n}Big)^n=lim_{nto+infty}Big(1+frac y{x+n}Big)^n=
    \lim_{nto+infty}frac{Big(1+frac y {x+n}Big)^{n+x}}{Big(1+frac y {x+n}Big)^x}=e^y
    $$
    Because $lim_{nto+infty}{Big(1+frac y {x+n}Big)^{n+x}}=e^y$ and $lim_{nto+infty}{Big(1+frac y {x+n}Big)^x}=1$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      $x+n$ is not necessarily a positive integer.
      $endgroup$
      – Paramanand Singh
      Nov 28 '18 at 1:52
















    1












    $begingroup$

    $$
    \dfrac{e^{x+y}}{e^x}=lim_{nto+infty}{Big(dfrac{1+frac {x+y} n}{1+frac x n}Big)^n}=
    \lim_{nto+infty}Big(frac{x+y+n}{x+n}Big)^n=lim_{nto+infty}Big(1+frac y{x+n}Big)^n=
    \lim_{nto+infty}frac{Big(1+frac y {x+n}Big)^{n+x}}{Big(1+frac y {x+n}Big)^x}=e^y
    $$
    Because $lim_{nto+infty}{Big(1+frac y {x+n}Big)^{n+x}}=e^y$ and $lim_{nto+infty}{Big(1+frac y {x+n}Big)^x}=1$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      $x+n$ is not necessarily a positive integer.
      $endgroup$
      – Paramanand Singh
      Nov 28 '18 at 1:52














    1












    1








    1





    $begingroup$

    $$
    \dfrac{e^{x+y}}{e^x}=lim_{nto+infty}{Big(dfrac{1+frac {x+y} n}{1+frac x n}Big)^n}=
    \lim_{nto+infty}Big(frac{x+y+n}{x+n}Big)^n=lim_{nto+infty}Big(1+frac y{x+n}Big)^n=
    \lim_{nto+infty}frac{Big(1+frac y {x+n}Big)^{n+x}}{Big(1+frac y {x+n}Big)^x}=e^y
    $$
    Because $lim_{nto+infty}{Big(1+frac y {x+n}Big)^{n+x}}=e^y$ and $lim_{nto+infty}{Big(1+frac y {x+n}Big)^x}=1$






    share|cite|improve this answer









    $endgroup$



    $$
    \dfrac{e^{x+y}}{e^x}=lim_{nto+infty}{Big(dfrac{1+frac {x+y} n}{1+frac x n}Big)^n}=
    \lim_{nto+infty}Big(frac{x+y+n}{x+n}Big)^n=lim_{nto+infty}Big(1+frac y{x+n}Big)^n=
    \lim_{nto+infty}frac{Big(1+frac y {x+n}Big)^{n+x}}{Big(1+frac y {x+n}Big)^x}=e^y
    $$
    Because $lim_{nto+infty}{Big(1+frac y {x+n}Big)^{n+x}}=e^y$ and $lim_{nto+infty}{Big(1+frac y {x+n}Big)^x}=1$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 27 '18 at 19:06









    Samvel SafaryanSamvel Safaryan

    493111




    493111












    • $begingroup$
      $x+n$ is not necessarily a positive integer.
      $endgroup$
      – Paramanand Singh
      Nov 28 '18 at 1:52


















    • $begingroup$
      $x+n$ is not necessarily a positive integer.
      $endgroup$
      – Paramanand Singh
      Nov 28 '18 at 1:52
















    $begingroup$
    $x+n$ is not necessarily a positive integer.
    $endgroup$
    – Paramanand Singh
    Nov 28 '18 at 1:52




    $begingroup$
    $x+n$ is not necessarily a positive integer.
    $endgroup$
    – Paramanand Singh
    Nov 28 '18 at 1:52











    2












    $begingroup$

    It suffices to show that
    $$
    lim_{ntoinfty}frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=1 tag{1}
    $$
    But
    $$
    frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=
    left(1+frac{xy}{n^2(1+frac{x+y}{n})}right)^n
    $$
    and as, for suitably large $n$, say $nge n_0$, we have that
    $$
    frac{1}{2}<1+frac{x+y}{n}<2,
    $$
    then
    $$
    left(1+frac{xy}{2n^2}right)^n<frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=
    left(1+frac{xy}{n^2(1+frac{x+y}{n})}right)^n<left(1+frac{2xy}{n^2}right)^n
    $$
    Now, for all $zinmathbb R$, we have
    $$
    left(1+frac{z}{n^2}right)^n=left(left(1+frac{z}{n^2}right)^{n^2}right)^{1/n}to 1,
    $$
    since $left(1+frac{z}{n^2}right)^{n^2}to e^z$.



    Thus
    $$
    lim_{ntoinfty}left(1+frac{xy}{2n^2}right)^n=lim_{ntoinfty}left(1+frac{2xy}{n^2}right)^n=1,
    $$
    and hence $(1)$ holds.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      It suffices to show that
      $$
      lim_{ntoinfty}frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=1 tag{1}
      $$
      But
      $$
      frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=
      left(1+frac{xy}{n^2(1+frac{x+y}{n})}right)^n
      $$
      and as, for suitably large $n$, say $nge n_0$, we have that
      $$
      frac{1}{2}<1+frac{x+y}{n}<2,
      $$
      then
      $$
      left(1+frac{xy}{2n^2}right)^n<frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=
      left(1+frac{xy}{n^2(1+frac{x+y}{n})}right)^n<left(1+frac{2xy}{n^2}right)^n
      $$
      Now, for all $zinmathbb R$, we have
      $$
      left(1+frac{z}{n^2}right)^n=left(left(1+frac{z}{n^2}right)^{n^2}right)^{1/n}to 1,
      $$
      since $left(1+frac{z}{n^2}right)^{n^2}to e^z$.



      Thus
      $$
      lim_{ntoinfty}left(1+frac{xy}{2n^2}right)^n=lim_{ntoinfty}left(1+frac{2xy}{n^2}right)^n=1,
      $$
      and hence $(1)$ holds.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        It suffices to show that
        $$
        lim_{ntoinfty}frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=1 tag{1}
        $$
        But
        $$
        frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=
        left(1+frac{xy}{n^2(1+frac{x+y}{n})}right)^n
        $$
        and as, for suitably large $n$, say $nge n_0$, we have that
        $$
        frac{1}{2}<1+frac{x+y}{n}<2,
        $$
        then
        $$
        left(1+frac{xy}{2n^2}right)^n<frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=
        left(1+frac{xy}{n^2(1+frac{x+y}{n})}right)^n<left(1+frac{2xy}{n^2}right)^n
        $$
        Now, for all $zinmathbb R$, we have
        $$
        left(1+frac{z}{n^2}right)^n=left(left(1+frac{z}{n^2}right)^{n^2}right)^{1/n}to 1,
        $$
        since $left(1+frac{z}{n^2}right)^{n^2}to e^z$.



        Thus
        $$
        lim_{ntoinfty}left(1+frac{xy}{2n^2}right)^n=lim_{ntoinfty}left(1+frac{2xy}{n^2}right)^n=1,
        $$
        and hence $(1)$ holds.






        share|cite|improve this answer









        $endgroup$



        It suffices to show that
        $$
        lim_{ntoinfty}frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=1 tag{1}
        $$
        But
        $$
        frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=
        left(1+frac{xy}{n^2(1+frac{x+y}{n})}right)^n
        $$
        and as, for suitably large $n$, say $nge n_0$, we have that
        $$
        frac{1}{2}<1+frac{x+y}{n}<2,
        $$
        then
        $$
        left(1+frac{xy}{2n^2}right)^n<frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=
        left(1+frac{xy}{n^2(1+frac{x+y}{n})}right)^n<left(1+frac{2xy}{n^2}right)^n
        $$
        Now, for all $zinmathbb R$, we have
        $$
        left(1+frac{z}{n^2}right)^n=left(left(1+frac{z}{n^2}right)^{n^2}right)^{1/n}to 1,
        $$
        since $left(1+frac{z}{n^2}right)^{n^2}to e^z$.



        Thus
        $$
        lim_{ntoinfty}left(1+frac{xy}{2n^2}right)^n=lim_{ntoinfty}left(1+frac{2xy}{n^2}right)^n=1,
        $$
        and hence $(1)$ holds.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 27 '18 at 19:10









        Yiorgos S. SmyrlisYiorgos S. Smyrlis

        63.2k1384163




        63.2k1384163






























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            ComboBox Display Member on multiple fields

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            2 I'm trying to set the .DisplayMember property of a ComboBox in C#, but I want to bind it to multiple columns in the .DataSouce . My SQL looks like this: SELECT PersNbr, PersFirstName, PersMiddleName, PersLastName FROM Pers WHERE PersNbr = :persNbr; I'm saving this query in a DataTable so each column selected has it's own column in the Datatable . I want to make the .DisplayMember a combination of PersFirstName + PersMiddleName + PersLastName so their full name appears like this: comboBox.DisplayMemeber = "PersFirstName" + "PersMiddleName" + "PersLastName" I know I can just to this on the query: SELECT PersNbr, (PersFirstName || PersMiddleName || PersLastName) PersName and then just do this: comboBox.DisplayMember = "PersName"; but I
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            Is it possible to collect Nectar points via Trainline?

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            2 Since I am not eligible for any Railcard, the only available discount/bonus scheme is Nectar Card for me. GWR, SWR, LNER and Virgin Trains allow to collect Nectar points and I have already linked a card to my GWR account. A friend suggested me to use Trainline App, because it can search for bus tickets either. I had a look at the application, but could not see any option for Nectar. Is there a way to collect Nectar points when I buy tickets via Trainline? uk trains loyalty-programs trainline share | improve this question asked Dec 2 '18 at 15:50 ahmedus 3,224 5 19 49
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