Solving a $1^\infty$ indeterminate form. I'm preparing for my calculus exam and I can't solve this limit:
$$\lim_{x\rightarrow\infty}\left(\frac{1+\tan(1/x)}{1-\tan(1/x)}\right)^x$$
The limit tends to $1^\infty$, which is indeterminate.
I've tried several things and I couldn't solve it.
Any idea? Thanks in advance.
 A: I have an alternative solution without the use of the L'Hospital rule. Start as Paul suggested, but when in the form of
$$
\lim_{x \to \infty} x \log \left(\frac{1+\tan(1/x)}{1-\tan(1/x)}\right)
$$
you can use the fact that
$$
\lim_{y \to 1} \frac{\log y}{y - 1} = 1.
$$
Using this limit, the limit arithmetic and a limit of a composed function. All that helps you transform the limit above into
$$
\lim_{x \to \infty} x \left(\frac{1+\tan(1/x)}{1-\tan(1/x)} - 1\right) = \lim_{x \to \infty} x \left(\frac{2\tan(1/x)}{1-\tan(1/x)}\right) = \lim_{x \to \infty} 2 \cdot \frac{\tan{1/x}}{\frac 1x}
$$
Going from the second part to the third one required yet another arithmetic to get rid of the denominator - that is obviously one, because it is continuous. The last bit can be solved using yet another known limit
$$
\lim_{y \to 0} \frac{\tan y}{y} = 1
$$
So we know the limit is two, we apply the exponential function and get the result $e^2$.
Hope this helps as well.
(Sorry for the typesetting mess [no eq numbers], I have yet to learn how to work with this system.)
A: Note that 
$$\tag{1}\lim_{x\rightarrow\infty}\left(\frac{1+\tan(1/x)}{1-\tan(1/x)}\right)^x=\lim_{x\rightarrow\infty}e^{\displaystyle x\ln\left(\frac{1+\tan(1/x)}{1-\tan(1/x)}\right)}=e^{\displaystyle\lim_{x\rightarrow\infty}x\ln\left(\frac{1+\tan(1/x)}{1-\tan(1/x)}\right)}$$
since $e^x$ is a continuous function. 
Note that 
$$\lim_{x\rightarrow\infty}x\ln\left(\frac{1+\tan(1/x)}{1-\tan(1/x)}\right)=
\lim_{x\rightarrow\infty}\frac{\ln\left(\frac{1+\tan(1/x)}{1-\tan(1/x)}\right)}{\frac{1}{x}}  \cdot \left(\frac{0}{0}\right)$$
We can apply the L'Hospital rule to the previous limit. Since
$$\frac{d}{dx}\ln\left(\frac{1+\tan(1/x)}{1-\tan(1/x)}\right)=\frac{1-\tan(1/x)}{1+\tan(1/x)}\cdot \frac{d}{dx}\left(\frac{1+\tan(1/x)}{1-\tan(1/x)}\right)
$$
$$=\frac{1-\tan(1/x)}{1+\tan(1/x)}\cdot \frac{d}{dx}\left(\frac{2}{1-\tan(1/x)}-1\right)=\frac{1-\tan(1/x)}{1+\tan(1/x)}\cdot \frac{2\sec^2(1/x)\cdot(-\frac{1}{x^2})}{(1-\tan(1/x))^2},$$
we have
$$\lim_{x\rightarrow\infty}\frac{\ln\left(\frac{1+\tan(1/x)}{1-\tan(1/x)}\right)}{\frac{1}{x}}=
\lim_{x\rightarrow\infty}\frac{\frac{1-\tan(1/x)}{1+\tan(1/x)}\cdot \frac{2\sec^2(1/x)\cdot(-\frac{1}{x^2})}{(1-\tan(1/x))^2}}{-\frac{1}{x^2}}$$
$$\tag{2}=\lim_{x\rightarrow\infty}\frac{1-\tan(1/x)}{1+\tan(1/x)}\cdot \frac{2\sec^2(1/x)}{(1-\tan(1/x))^2}=2.$$
Combining $(1)$ and $(2)$, we have 
$\lim_{x\rightarrow\infty}\left(\frac{1+\tan(1/x)}{1-\tan(1/x)}\right)^x=e^2.$
A: Asymptotics ...
$$\begin{align}
\operatorname{tan} \biggl(\frac{1}{x}\biggr) &= \frac{1}{x} + \frac{1}{3 x^{3}} + \frac{2}{15 x^{5}} + O \Bigl(x^{(-6)}\Bigr)\\
1 + \operatorname{tan} \biggl(\frac{1}{x}\biggr) &= 1 + \frac{1}{x} + \frac{1}{3 x^{3}} + \frac{2}{15 x^{5}} + O \Bigl(x^{(-6)}\Bigr)\\
1 - \operatorname{tan} \biggl(\frac{1}{x}\biggr) &= 1 - \frac{1}{x} - \frac{1}{3 x^{3}} - \frac{2}{15 x^{5}} + O \Bigl(x^{(-6)}\Bigr)\\\frac{1 + \operatorname{tan} \Bigl(\frac{1}{x}\Bigr)}{1 - \operatorname{tan} \Bigl(\frac{1}{x}\Bigr)} &= 1 + \frac{2}{x} + \frac{2}{x^{2}} + \frac{8}{3 x^{3}} + \frac{10}{3 x^{4}} + \frac{64}{15 x^{5}} + O \Bigl(x^{(-6)}\Bigr)\\\left(\frac{1 + \operatorname{tan} \Bigl(\frac{1}{x}\Bigr)}{1 - \operatorname{tan} \Bigl(\frac{1}{x}\Bigr)}\right)^{x} &= \operatorname{e} ^{2} + \frac{4 \operatorname{e} ^{2}}{3 x^{2}} + \frac{20 \operatorname{e} ^{2}}{9 x^{4}} + O \Bigl(x^{(-5)}\Bigr)
\end{align}$$
A: You may let the limit as $z$ and let $y=\ln(z)$, then use L'Hospital rule to find the limits of $y$ and finally $z$ can be calculated $\exp (y)$
A: EDIT:
you can write your expression as 
$$
\bigg(1+\frac{2\tan(1/x)}{1-\tan(1/x)}\bigg)^x \sim \bigg(1+\frac{2}{x}\bigg)^x \rightarrow e^2
$$
when $x\rightarrow \infty$.
