Evaluate the Limit: $\lim_{x\to \pi/2^-} (\tan x)^{\cos x}$ $$\lim_{x\to (\pi/2)^-} (\tan x)^{\cos x}$$
I am supposed to use $\ln$ but I am not sure as to why since I thought I used $\ln$ when there is variable as the base and the exponent. I do not see this in this case. 
Can someone explain?  
 A: $$\ln\bigl(\tan x^{\cos x}\bigr)=\cos x\ln(\sin x)-\cos x\ln(\cos x)$$
The first term tends to $0\ln 1=0$ by continuity. The second term tends to $0$ since $\lim\limits _{u\to 0_+}(u\ln u)=0$.
A: $$
\lim_{x\to (\pi/2)^-} (\tan x)^{\cos x} = \frac 1 {\lim\limits_{x\to (\pi/2)^-}(\cot x)^{\cos x}}
$$
Notice that $\cot x$ and $\cos x$ both cross the $x$-axis at a $45^\circ$ angle at $\pi/2$.  That tells you that the limit in the denominator is essentially $\lim\limits_{x\,\downarrow\,0} x^x$, which is $1$. To show that that is $1$, use L'Hopital's rule, thus:
\begin{align}
& \lim\limits_{x\,\downarrow\,0} x^x = \exp\left( \lim\limits_{x\,\downarrow\,0} x\ln x \right) = \exp\left( \lim\limits_{x\,\downarrow\,0} \frac{\ln x}{1/x} \right) \\[10pt]
= {} & \exp\left( \lim_{x\,\downarrow\,0} \frac{1/x}{-1/x^2} \right) = \exp \left(\lim_{x\,\downarrow 0} -x \right) = \exp 0 = 1.
\end{align}
(This still leaves the issue of actually proving that if $f(x)$ and $g(x)$ both have slope $1$ when they cross the axis, then $\lim f(x)^{g(x)}= \lim x^x$.
A: $$\lim_\limits{x\to (\pi/2)^-} (\tan x)^{\cos x}=\lim_\limits{x\to (\pi/2)^-} e^{{\cos x}\ln(\tan x)}=e^{\lim_\limits{x\to (\pi/2)^-}{{\cos x}\ln(\tan x)}}=e^{\lim_\limits{x\to (\pi/2)^-}{{1\over {1\over \cos x}}\ln(\tan x)}}.$$ 
Now, $\tan ({\pi\over 2})$ is not defined. However, $\tan (x)$ tends to infinity as $x$ tends to $\pi \over 2$ from below. At the same time, $1\over \cos x$ tends to infinity as well, as $\cos x$ tends to zero in that tendency of $x$. We, therefore, can apply L'Hôpital's rule: $${\lim_\limits{x\to (\pi/2)^-}{{1\over {1\over \cos x}}\ln(\tan x)}}={\lim_\limits{x\to (\pi/2)^-}{{1\over {\sin x\over \cos^2 x}}\cdot{1\over \tan x}}}\cdot{1\over \cos ^2 x}=\lim_\limits{x\to (\pi/2)^-}{{1\over \sin x\tan x}}=\lim_\limits{x\to (\pi/2)^-}{{\cos x\over \sin^ 2 x}}={0\over 1}=0.$$  Having already used the continuity of $e^x$ when moving the limit elsewhere and computing that limit,you can now go back and place the result. 
A: HINT:consider $e^{\ln(\tan(x))\cos(x)}$ and use the rules of L'Hospital
A: Here's a slightly different approach from the others.  We will rely on only the Squeeze Theorem along with the elementary inequalities from geometry
$$x\cos x\le \sin x\le x \tag 1$$
for $\pi/2>x>0$.  
From $(1)$ it is straightforward to show that for $\pi/2>x>0$
$$\frac{\cos x}{x}\le \cot x\le \frac1x \tag 2$$
Now, we enforce the substitution $x -\pi/2 \mapsto x$.  Then, the limit of interest is 
$$\lim_{x\to \pi/2^-}(\tan x)^{\cos x}=\lim_{x\to 0^+}(\cot x)^{\sin x}$$
Using the inequalities in $(1)$ and $(2)$ we have for $\cos x\ge x$
$$\left(\frac{\cos x}{x}\right)^{x\cos x}\le(\cot x)^{\sin x}\le \left(\frac1x\right)^{x}$$
We can rewrite $(3)$ as
$$\left(\frac{1}{\left(\frac{x}{\cos x}\right)^{x/\cos x}}\right)^{\cos^2x}\le (\cot x)^{\sin x}\le \left(\frac1{x^x}\right) \tag 4$$
Recalling the $\lim_{x\to 0^+}x^x=1$, observing that as $x\to 0^+$, $x/\cos x \to 0^+$ also, and using the Squeeze theorem reveals
$$\lim_{x\to 0+}(\cot x)^{\sin x}=1$$
Therefore, we arrive at the limit of interest
$$\lim_{x\to \pi/2^-}(\tan x)^{\cos x}=1$$
