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?
 A: $\lim\limits_{x \rightarrow 0}\exp (x\ln (1+x))=\exp(\lim\limits_{x \rightarrow 0}(x\ln(1+x)))=\exp(0)=1$. 
$\lim\limits_{ x \rightarrow 0}\exp ( \frac{\ln (1+x)}{x})=\exp(\lim\limits_{x \rightarrow 0}(\frac{\ln(1+x)}{x}))=\exp(1)=e$. Use L'Hospital.. to see $\lim\limits_{x \rightarrow 0}\frac{\ln(1+x)}{x}=1$
A: 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 $\ln(L)$.
A: 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
$$
\log\left(\lim_{x\to0}(1+x)^{1/x}\right)=\lim_{x\to0}\frac{\log(1+x)}{x}
$$
A: 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
$$\lim_{n\rightarrow \infty}\left( 1+\frac{1}{n}\right) ^{n}=e.$$
A: Sorry for uploading the image - i am new and have yet to figure out how to mark up the math
The first line assumes you know that if f(x) = ln(x) then f'(1) = (ln(x+h) - ln(x)) / h
ps f'(1) = 1

