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. 

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 $0$.
