Algebraic manipulations for a simple limit 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=$$\lim\limits_{x\rightarrow\infty}(1+\frac{1}{x})^{xa}$. What are these algebraic operations? I apologize in advance if maybe this question is a little too simple, but I can't for the life of me figure it out. Thanks.
 A: Let's denote:
$A=(1+\frac{a}{x})^x$  , if we aplly logarithm on both sides we get:
$\ln A=x\ln(1+\frac{a}{x})$ , now make substitution $t=\frac{1}{x}$ , so we may write:
$\ln A=\frac{\ln (1+at)}{t}\Rightarrow$ $$\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} $$
A: *

*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): $$\large\frac{1}{\left(1+\frac{1}{x}\right)^{-a(x+1)}} .$$

A: 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.
