Evaluating $\lim\limits_{x\to 0}(1+x)^{1/x}$?

How would I work out a limit of the form:

$$\lim_{x\to 0}\;(1+x)^{1/x}$$

I know these types of limits have a solution based on $e$ but how do I find this solution?

$\lim\limits_{x \rightarrow 0}\exp (x\ln (1+x))=\exp(\lim\limits_{x \rightarrow 0}(x\ln(1+x)))=\exp(0)=1$.

$\lim\limits_{ x \rightarrow 0}\exp ( \frac{\ln (1+x)}{x})=\exp(\lim\limits_{x \rightarrow 0}(\frac{\ln(1+x)}{x}))=\exp(1)=e$. Use L'Hospital.. to see $\lim\limits_{x \rightarrow 0}\frac{\ln(1+x)}{x}=1$

Hint:

The functions $y = \log x$ and $y = e^x$ are continuous, and continuous functions respect limits: $$ \lim_{n \to \infty} f(g(n)) = f\left( \lim_{n \to \infty} g(n) \right), $$ for all continuous functions $f$, whenever $\displaystyle\lim_{n \to \infty} g(n)$ exists. Let $$L=\lim\limits_{x\to 0}(1+x)^{1/x}$$ be the limit which you to wish to find. Instead of finding $L$ directly, try on your own to find $\ln(L)$.

Try writing this as $$ \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n $$ The binomial theorem may be of some help then.

Another way of looking at this is taking logs and using L'Hospital $$ \log\left(\lim_{x\to0}(1+x)^{1/x}\right)=\lim_{x\to0}\frac{\log(1+x)}{x} $$

If we use the substitution $x=\frac{1}{y}$, since $\lim_{x\rightarrow 0}x=\lim_{y\rightarrow \infty }\frac{1}{y}$, we get

$$\lim_{x\rightarrow 0}\left( 1+x\right) ^{1/x}=\lim_{y\rightarrow \infty }\left( 1+\frac{1}{y}\right) ^{y}=e,$$

which uses the result

$$\lim_{n\rightarrow \infty}\left( 1+\frac{1}{n}\right) ^{n}=e.$$