How to evaluate $\lim_{x\to 0} (1+2x)^{1/x}$ Good night guys! I'm having some trouble with this: 
$$\lim_{x\to 0} (1+2x)^{1/x}$$
I know that $\lim_{x\to\infty} (1 + 1/x)^x = e$
but I don't know if i should take $h=1/(2x)$ or $h=1/x$
Can someone please help me? Thanks! 
 A: We first find the limit as $x$ approaches $0$ from the right.
Let $y=\frac{1}{2x}$. Then $2x=\frac{1}{y}$ and $\frac{1}{x}=2y$. We want
$$\lim_{y\to\infty}\left(1+\frac{1}{y}\right)^{2y}.$$
This is the square of the familiar 
$$\lim_{y\to\infty}\left(1+\frac{1}{y}\right)^{y}.$$
We conclude that
$$\lim_{x\to 0^+}\left(1+2x\right)^{1/x}=e^2.$$
Limit from the left is more troublesome with this approach. Let $1+2x=\frac{t}{1+t}$. Then $\frac{1}{x}=-2\frac{1+t}{t}$. So we are looking for 
$$\lim_{t\to 0^+}\frac{1}{\left(1+t\right)^{-2(1+t)/t}},$$
which is 
$$ \lim_{t\to 0^+}\left(1+t\right)^{2(1+t)/t}.$$
Now an argument much like the one before works.   The limit from the left is also $e^2$. 
A: For problems like these, try taking the natural log of the expression. More precisely, compute $\ln(\lim_{x\rightarrow 0} (1+2x)^{1/x}) = \lim_{x\rightarrow 0} \ln((1+2x)^{1/x}) = \lim_{x\rightarrow 0} \frac{\ln((1+2x))}{x}$. Now it is a form that allows you to apply L'Hopital. Don't for get to exponentiate at the end!
A: $$\displaystyle\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x=e,\displaystyle\lim_{x\to-\infty}\left(x+\frac{1}{x}\right)^x=e$$
$(-\infty,0),(0,\infty)\subset\mathbb R-\{0\}=\text{domain}\displaystyle\left(1+\frac{1}{x}\right)^x.$ So $\displaystyle\lim_{x\to0+}(1+x)^\frac{1}{x}=\displaystyle\lim_{x\to0-}(1+x)^\frac{1}{x}=e.$
Now $0$ is a limit point of both $(-\infty,0)$ and $(0,\infty)\implies\displaystyle\lim_{x\to0}(1+x)^\frac{1}{x}=e.$ Let 
$$f:\mathbb R-\{0\}\to\mathbb R:x\mapsto2x,\\g:\mathbb R-\{0\}\to\mathbb R:x\mapsto(1+x)^\frac{1}{x}$$
$f(\mathbb R-\{0\})\subset\mathbb R-\{0\}\implies gf:\mathbb R-\{0\}\to\mathbb R:x\mapsto(1+2x)^\frac{1}{2x}$ is defined and $\displaystyle\lim_{x\to0}f(x)=0$ being a limit point of $\mathbb R-\{0\}$ and $\displaystyle\lim_{x\to0}g(x)=e,$ we have $\displaystyle\lim_{x\to0}(1+2x)^\frac{1}{2x}=e.$
Consequently, $\displaystyle\lim_{x\to0}(1+2x)^\frac{1}{2x}=\displaystyle\lim_{x\to0}\left((1+2x)^\frac{1}{2x}\right)^2=e^2.$
