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I know what solution is $e^a$ but I don't know how to calculate this limit:

$$\lim \limits_{n \to \infty}\left(1 + \frac{a}{n}\right)^n=e^a$$

Can someone explain the steps to me?

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What definition do you know for $e$? Do you know the value of $\lim (1+\frac{1}{n})^n$? –  Jon Sep 12 '12 at 13:50

4 Answers 4

up vote 1 down vote accepted

Use $\lim_{n\rightarrow\infty} (1+1/n)^n=e$.

Let $m=n/a$ then $\lim_{n\rightarrow\infty} (1+a/n)^n=\lim_{m\rightarrow \infty} (1+1/m)^{ma}=e^a$.

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This proof is not self-contained: the variable $m$ is real. You should play with the integer part of $m$. –  Siminore Sep 12 '12 at 14:00
    
I think kspacja wants a proof which doesn't rely on using $\lim (1+1/n)^n=e$ –  user39572 Sep 12 '12 at 14:09
    
There may remain some work to do if $a\le 0$ is allowed. –  André Nicolas Sep 12 '12 at 15:23

$$\lim \limits_{x \to +\infty} (1+a/x)^x = e^{\lim \limits_{x \to +\infty} x \log (1+a/x)}$$

Then apply l'Hôpital's rule to $\frac{\log (1+a/x)}{1/x}$ to get $$\frac{\frac{1}{1+a/x} \cdot \frac{-a}{x^2}}{\frac{-1}{x^2}} = \frac{a}{1+a/x} = a$$

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Let $y=1/n\implies \lim \limits_{n \to \infty}\left(1 + \frac{a}{n}\right)^n=\lim_{y\to 0^+}(1+ay)^{\frac{1}{y}}$

Let $$A=\lim_{y\to 0^+}(1+ay)^{\frac{1}{y}}$$ $$\implies \log A=\lim_{y\to 0^+}\frac{\log(1+ay)}{y}=\lim_{y\to 0^+} \frac{a}{(1+ay)}=a$$ (applying L'Hopital's Rule) $$\implies \log A=a\implies A= e^a$$

Thus, $$\lim \limits_{n \to \infty}\left(1 + \frac{a}{n}\right)^n=e^a$$

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Of course, this approach requires a lot of machinery that may not yet be available. In particular, it is assumed that exponentiation with real exponents is defined and known to be continuous, and that it is known that the natural logarithm is continuous as well. Oh, and the limit requivalence in the second step is non-trivial as well. –  Johannes Kloos Sep 12 '12 at 14:15

Here is a more tedious way:

Let $c_{n,k} = \begin{cases} 0 && k>n \\ \binom{n}{k} \frac{1}{n^k} && \text{otherwise} \end{cases}$

Note that $0 \leq c_{n,k} \leq \frac{1}{k!}$, $c_{n+1,k} \geq c_{n,k}$, and $\lim_n c_{n,k} =\frac{1}{k!}$. (Showing these facts is the tedious part.) And, of course, $(1+\frac{a}{n})^n = \sum_{k=0}^\infty c_{n,k} a^k$.

Now note that $|e^a - (1+\frac{a}{n})^n| \leq \sum_{k=0}^\infty |\frac{1}{k!}-c_{n,k}| |a^k| \leq \sum_{k=0}^\infty \frac{1}{k!} |a^k| \leq e^{|a|}$. (The latter uses the fact that $0 \leq \frac{1}{k!}-c_{n,k} \leq \frac{1}{k!}$.)

Let $\epsilon>0$ and choose $N$ such that $\sum_{k>N} \frac{1}{k!} |a^k| < \frac{\epsilon}{2}$. Let $M = \max(1, |a|,...,|a|^N)$. Now choose $N' \geq N$ such that $\max_{k=0,...,N} |\frac{1}{k!}-c_{n,k}| < \frac{1}{M(N+1)} \frac{\epsilon}{2}$ whenever $n>N'$.

Then we have $|e^a - (1+\frac{a}{n})^n| \leq \sum_{k=0}^\infty |\frac{1}{k!}-c_{n,k}| |a^k| \leq \sum_{k=0}^N |\frac{1}{k!}-c_{n,k}| |a^k|+ \sum_{k>N} |\frac{1}{k!}-c_{n,k}| |a^k|$. Both terms on the right hand side are bounded by $\frac{\epsilon}{2} $, hence we have $|e^a - (1+\frac{a}{n})^n| < \epsilon$, for $n> N'$.

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