Calculate the limit $\lim_{h\to 0} (1+ 2h)^\frac{1}{h}$ So I was working on it and I got that
$\lim_{h\to 0} (1+ 2h)^\frac{1}{h}= 1$. 
But then I searched on Wolfram and I got that
$\lim_{h\to 0} (1+ 2h)^\frac{1}{h}={e^2}  $. 
So, is my answer right? or how should I get the same result as Wolfram Alpha?
 A: Recall that $e^x= \lim_{n \to \infty} (1 + x/n)^n$. 
Now, in your limit replace $1/h$ by $n$.
A: $$\large\lim_{h\to 0} (1+ 2h)^\frac{1}{h}$$
$$\large= \lim_{h\to 0} e^{ln{(1+ 2h)^\frac{1}{h}}}$$
$$\large= \lim_{h\to 0} e^{\frac{ln{(1+ 2h)}}{h}}$$
$$\large=e^{\lim\limits_{h\to 0}\frac{ln{(1+ 2h)}}{h}}$$
$$=e^2$$
$$\large\lim_{x\to 0} \frac{ln{(1+ ax)}}{x}=a$$
A: $$ \lim\limits_{h\to 0} (1+ 2h)^{\frac{1}{h}} $$
$$ = \lim\limits_{h\to 0} \exp\left(\ln\left(1+ 2h\right)^{\frac{1}{h}}\right) $$
$$ = \lim\limits_{h\to 0} \exp\left(\frac{\ln\left(1+ 2h\right)}{h}\right) $$
$$ = \exp\left(\lim\limits_{h\to 0}\frac{\ln\left(1+ 2h\right)}{h}\right) $$
$$ = \exp\left(\lim\limits_{h\to 0}\frac{\frac{d}{dh}\ln\left(1+ 2h\right)}{\frac{d}{dh}h}\right) $$
$$ = \exp\left(\lim\limits_{h\to 0}\frac{2}{1+2h}\right) $$
$$ = \exp\left(2\right) $$
A: Consider $$A=(1+ 2h)^\frac{1}{h}$$ Take logarithms of both sides $$\log(A)=\frac{1}{h}\log(1+2h)$$ Now, using that for small $x$, $\log(1+x)=\frac{x}{1}-\frac{x^2}{2}+\frac{x^3}{3}+\cdots$. Replace $x$ by $2h$ so $$\log(A)=\frac{1}{h}\Big(2 h-2 h^2+\frac{8 h^3}{3}+\cdots \Big)=2-2h+\frac{8 h^2}{3}$$ The limit is then clearly $\log(A)=2$ that is to say $A=e^2$.
But we can do a bit more and get more information about what is going on around $h=0$. Rewrite $$A=e^2 e^{-2h+\frac{8 h^2}{3}}$$ and using that for small $x$, $e^x=1+\frac{x}{1}+\frac{x^2}{2}+\cdots$, replace $x$ by $-2h+\frac{8 h^2}{3}$ to get $$A=e^2\Big(1-2 h+\frac{14 h^2}{3}+\cdots\Big)$$ which gives the limit and also shows how the limit is approached.
Making the problem more complex, considering $$B=(1+ ah)^\frac{b}{h}$$ the same procedure would lead to $$B=e^{ab}\Big(1-\frac{a^2b}{2} h +\frac{a^3 b(3 a b+8)}{24}  h^2 +\cdots\Big)$$
