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I can't seem to prove that $e^a=\lim\limits_{x\rightarrow\infty}(1+\frac{a}{x})^x$. I'm sure there must be some algebraic manipulations that can be done in order to show that $\lim\limits_{x\rightarrow\infty}(1+\frac{a}{x})^x=$$\lim\limits_{x\rightarrow\infty}(1+\frac{1}{x})^{xa}$. What are these algebraic operations? I apologize in advance if maybe this question is a little too simple, but I can't for the life of me figure it out. Thanks.

Hautdesert
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  • $\lim_{x\to \infty} {\left({\frac{a}{x}+1}\right)}^x$ can also be written as $e^{\lim_{x \to \infty} x \log {(\frac{a}{x}+1)}}$ Now, the as the limit equals to $a$, the expression equals to $e^{a}$. – Gaurav Tiwari Oct 11 '11 at 04:15
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    The proof depends on the precise definitions that one makes. For many introductions to calculus, you would start by taking the natural logarithm of $(1+a/x)^x$. But if $e$ is defined as the limit of $(1+1/y)^y$ as $y \to \infty$, you would, for positive $a$, note that $(1+a/x)^x=((1+a/x)^{x/a})^a$. As $x\to\infty$, $x/a \to \infty$, so $(1+a/x)^{x/a} \to e$. – André Nicolas Oct 11 '11 at 04:20
  • In short: what definition of the exponential function are you using? Your limit is actually one definition used for the exponential function... – J. M. ain't a mathematician Oct 11 '11 at 04:29
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    As @J.M. and André say there are several definitions of $e$ or the exponential function you could be using. You might want to look at Prove the definitions of $e$ to be equivalent or Combinatorial proof for some different approaches. – robjohn Oct 11 '11 at 05:31
  • @Hautdesert: In connection with some of the other comments here (e.g. it depends on how $e$ is defined), some or all of the various manipulations that I recently posted in the ap-calculus listserv (use the URL that follows) could be of use in your class. http://mathforum.org/kb/message.jspa?messageID=7575015 – Dave L. Renfro Oct 11 '11 at 14:41

3 Answers3

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Let's denote:

$A=(1+\frac{a}{x})^x$ , if we aplly logarithm on both sides we get:

$\ln A=x\ln(1+\frac{a}{x})$ , now make substitution $t=\frac{1}{x}$ , so we may write:

$\ln A=\frac{\ln (1+at)}{t}\Rightarrow$ $$\lim\limits_{x\rightarrow\infty} A = e^{\lim\limits_{t\rightarrow 0} (\frac{\ln (1+at)}{t})} $$

Now if we apply L'Hopital rule on this limit we can write:

$$\lim\limits_{x\rightarrow\infty} A=e^{\lim\limits_{t\rightarrow 0}(\frac{a}{at+1})} =e^{a} $$

Pedja
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  • For $a>0$, you can replace $x$ with $ax$ in the expression: as $x\to\infty$, so does $ax$.
  • For $a=0$, the limit is trivial.
  • For $a<0$, you can flip the expression upside down as $$\large\frac{1}{\left(1-\frac{a}{x+a}\right)^x},$$ then replace $x$ with $-a(x+1)$ (it goes to $\infty$ just the same as well): $$\large\frac{1}{\left(1+\frac{1}{x}\right)^{-a(x+1)}} .$$
anon
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As André Nicolas comments, often this equation is used to define $e$, for which case he has given a proof. Assuming instead that $e^x := \lim_{n\to\infty}\sum_{i=0}^{n} \frac{x^i}{i!}$, expand the binomial and show that the corresponding terms have equal coefficients.

Dan Brumleve
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