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.
Algebraic manipulations for a simple limit
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real-analysis
limits
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0@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 – 2011-10-11
3 Answers
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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} $
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0What definition of $e$ or $e^x$ is being implicitly used here? I'm not sure that it doesn't assume the conclusion. – 2011-10-11
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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): $\large\frac{1}{\left(1+\frac{1}{x}\right)^{-a(x+1)}} .$
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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.