calculus 2 limit question

Question

Prove that $$\lim_{x\to\infty}\left(1+\frac{a}{x}\right)^x = e^a$$ for $a$ not equal to $0$.

Answer 1

Start with the case of $a > 0$.

Start from one of the definitions of $e$ (probably the only one you started with in your class: $$ \lim_{n\to\infty}\left( 1 + \frac{1}{n} \right)^n = e $$ Now consider a sequence $S_n$ each of whose $n$-th term is $$ \left( left( 1 + \frac{1}{n} \right)^n \right) ^a = left( 1 + \frac{1}{n} \right)^{na} $$ It is easy to show (using the $\delta-\epsilon$ definition of limits), that $$ \lim_{n\to\infty} S_n = e^a$$ For choose some $\epsilon > 0$. Then (taking the test $\varepsilon$ in the limit definition to be $\epsilon^{1/a}$, since $\lim_{n\to\infty}\left( (S_n)^{1/a} = e $, there exists some finite $\delta$ such that $$n>\delta \rightarrow |(S_n)^{1/a} - e| < \varepsilon = \epsilon^{1/a} $$ so whenever $n>\delta$, $$ |S_n - e^a| < \epsilon $$

SO we have establisshed that $$ \lim_{n\to\infty}\left( 1 + \frac{1}{n} \right)^{na} = e^a $$ Now let $x = na$; when $n$ goes to infinity, so does $x$. Now $n = a/x$ so $$ \lim_{n\to\infty}\left( 1 + \frac{a}{x} \right)^{x} = e^a $$

The $a < 0$ case goes along similar lines.

Since each term