Find $\lim_{x \to 0}(1+\frac x m)^\frac 1 x$ where $m\ne 0$. 
Find $\displaystyle \lim_{x \to 0}(1+\frac x m)^\frac 1 x$ where $m\ne 0$.

I know that I've to use $\lim_{x\to 0} \frac{e^x-1} x=1$, but I don't know how. Please help. Thank you.
 A: This is my way :
 $$\lim_{x\to 0}(1+\frac{x}{m})^{\frac{1}{x}}$$ 
$$=\lim_{x\to 0}(1+\frac{1}{\frac{m}{x}})^{\frac{m}{x}.\frac{1}{m}}$$
$$=e^\frac{1}{m}$$
A: This is a list of hints.
$1.$ Try $x\mapsto um$. Then you'll get something depending on $u$ to the power of $m^{-1}$.
$2.$ You know $$\frac{e^x-1}x\to 1$$
Use $e^x-1\mapsto u$. Then you get that $$\frac{u}{\log(1+u)}\to 1$$
Use that $$r\log x=\log(x^r)$$ for any real $r$ to get something similar to what you have in $1.$. You'll have to use continuity of the logarithm to "bring the limit inside".
Spoilers ahead
$1.$ You should get to $$=\lim\limits_{u\to 0}\left(1+u\right)^{\frac 1 u\frac 1 m}$$
$2.$ ou get $$ \begin{align} 1 &=& \lim\limits_{u\to 0}\frac{u}{\log(1+u)}\\&=&\lim\limits_{u\to 0}\frac{1}{\frac{\log(1+u)}u}\\&=& \lim\limits_{u\to 0}\frac{1}{{\log(1+u)^{1/u}}}\\&=& \frac{1}{\lim\limits_{u\to 0}{\log(1+u)^{1/u}}}\end{align}$$ That is $$1={\lim\limits_{u\to 0}{\log(1+u)^{1/u}}}$$ 
Because the logarithm is continuous over the positive numbers, and that limit is positive, we can say 
$$1={{\log \lim\limits_{u\to 0}(1+u)^{1/u}}}$$ 
What does this tell you about you original limit?
A: $$\left(1+\frac{x}{m}\right)^{1/x}=\left[\left(1+\frac{1}{\frac{m}{x}}\right)^{m/x}\right]^{1/m}\xrightarrow[x\to 0]{}e^{1/m}$$
Since
$$\left(1+\frac{1}{f(x)}\right)^{f(x)}\xrightarrow[x\to x_0]{}e$$
whenever $\,f(x)\xrightarrow[x\to x_0]{}\infty\,$
A: This limit becomes easy if you know $\log(1+x)\sim x$ if $x$ is close to $0$.
So let's use this 
$$\lim_{x\to 0}\left(1+\frac{x}{m}\right)^{\frac{1}{x}}=\lim_{x\to 0}e^{\frac{1}{x}\log\left(1+\frac{x}{m}\right)}=\lim_{x\to 0}e^{\frac{1}{x}\frac{x}{m}}=e^{\frac{1}{m}}.$$
