Why is the limit of ${\tan(x)}^{\cos( x)}$ as $x$ approaches $\pi/2$ equals 1? Why is the limit of ${\tan(x)}^{\cos( x)}$ as $x$ approaches $\pi/2$ equals 1? Please explain and include step by step solution. 
I've changed it into the lny = cos ln tan x. That is equal to infinity X 0. What would be the next step. 
 A: Whenever $x\in(0,\pi/2)$, $$\tan(x)^{\cos(x)}=\exp[\cos(x)\log(\tan(x))].$$ As $x\nearrow\pi/2$, $\tan(x)\to+\infty$, so that $\log(\tan(x))\to\infty$. On the other hand, $\cos(x)$ converges to $0$, so that this limit is of the form “$0\cdot\infty$”.
Let's try converting it to a fraction and use L'Hôpital's rule:
\begin{align*}
&\,\lim_{x\nearrow\pi/2}\cos(x)\log(\tan(x))=\lim_{x\nearrow\pi/2}\frac{\log(\tan(x))}{\dfrac{1}{\cos(x)}}=\lim_{x\nearrow\pi/2}\frac{\dfrac{1}{\tan(x)}\cdot\dfrac{1}{\cos^2(x)}}{-\dfrac{1}{\cos^2(x)}\cdot(-\sin(x))}\\
=&\,\lim_{x\nearrow\pi/2}\frac{\cos(x)}{\sin^2(x)}=0.
\end{align*}
Now,
\begin{align*}
&\,\lim_{x\nearrow\pi/2}\tan(x)^{\cos(x)}=\lim_{x\nearrow\pi/2}\exp[\cos(x)\log(\tan(x))]=\exp\left[\lim_{x\nearrow\pi/2}\cos(x)\log(\tan(x))\right]\\
=&\,\exp(0)=1.\end{align*}
Note that this is the limit from the left. The right-hand side limit does not exist, because $\tan(x)<0$ for $x\in(\pi/2,\pi/2+\varepsilon)$ for $\varepsilon>0$ small enough, and, in general, you can't raise a negative number to an arbitrary power (only in special cases). In fact, for such values, the function is not even defined.
A: For $x$ approaching $\pi/2$ from below:
Instead of using L'Hospital, one can also write
$$\cos(x)\log(\tan x ) = \cos(x)\log(\frac{\sin x }{\cos x}) = \cos(x)\log(\sin x) - \cos(x)\log(\cos x) $$
where each of the two terms on the right tends to $0$ separately -- the first one simply by continuity, the second one by the substitution $u=\cos x$, which makes the limit into $\lim_{u\to 0^{+}}u \log u$ which is well known to be $0$.
