How do I evaluate this limit: $\lim_{x\rightarrow\pm\infty}\frac{e^{2/x}-1}{\ln\frac{2x+1}{2x}}$? $$\lim_{x\rightarrow\pm\infty}\frac{e^{2/x}-1}{\ln\frac{2x+1}{2x}}$$
To me, this just seems impossible. I'm trying to use one of the common limits to figure this out, but I have absolutely no clue on where to start.
 A: Using as weak tools, as it is possible:
$$
\lim_{x\rightarrow\pm\infty}\frac{e^{2/x}-1}{\ln\frac{2x+1}{2x}}=
\lim_{x\rightarrow\pm\infty}\frac{\frac{e^{2/x}-1}{\frac2x}}{\frac{\ln(1+\frac1{2x})}{\frac1{2x}}}\cdot\frac{\frac2x}{\frac1{2x}}=4.
$$
(You should know limits $\lim_{y\to0}\frac{\ln(1+y)}{y}=1$ and $\lim_{y\to0}\frac{e^y-1}{y}=1$).
A: For me it is useful to let $t=1/x$. 
Then the numerator is $e^{2t}-1$ and the denominator is $\ln(1+t/2)$.
We are interested in the limit as $t$ approaches $0$ from the right, and from the left,  of
$$\frac{e^{2t}-1}{\ln(1+t/2)}.$$
Many things work, such as L'Hospital's Rule, or series expansion.
A: First it helps to identify which indeterminate form you are in. If you are not already in indeterminate form, you have to arrange your quantity so that it is. Fortunately, your problem is already of the form $\frac{0}{0}$ so we can apply L'Hospital's rule. L'Hospital's rule tells us that $$\lim_{x\rightarrow\pm\infty}\frac{e^{2/x}-1}{\ln\left(1+\frac{1}{2x}\right)} = \lim_{x\rightarrow\pm\infty}\frac{e^{2/x}\cdot \frac{-2}{x^2}}{\frac{1}{\left(1+\frac{1}{2x}\right)}\cdot \frac{-1}{2x^2}}$$ Now if you do a bit of algebra you should find that $$\frac{e^{2/x}\cdot \frac{-2}{x^2}}{\frac{1}{\left(1+\frac{1}{2x}\right)}\cdot \frac{-1}{2x^2}} = \frac{-2e^{2/x}}{\frac{2x}{2x+1}\cdot \frac{-1}{2x^2} \cdot x^2} \\= \frac{2e^{2/x}}{\frac{x}{2x+1}} \\ = \frac{2(2x+1)e^{2/x}}{x} \\ = 4e^{2/x}+\frac{2e^{2/x}}{x}$$ and hence  $$\lim_{x\rightarrow\pm\infty}\frac{e^{2/x}-1}{\ln\left(1+\frac{1}{2x}\right)} = \lim_{x\rightarrow\pm\infty}4e^{2/x}+\frac{2e^{2/x}}{x} \\ = \lim_{x\rightarrow\pm\infty}4e^{2/x}+\lim_{x\rightarrow\pm\infty}\frac{2e^{2/x}}{x} \\ = 4 + 0 \\ = 4$$
A: using L'Hosptal we obtain 
$-{\frac {2\,x+1}{{x}^{3}}{{\rm e}^{2\,{x}^{-1}}} \left( {x}^{-1}-1/2\,
{\frac {2\,x+1}{{x}^{2}}} \right) ^{-1}}
$
simplifying this we get
$2\,{\frac {2\,x+1}{x}{{\rm e}^{2\,{x}^{-1}}}}$ 
