How to find the limit of trig functions in exponents? In my Calculus course, I am studying exponential functions and their involvement in limits.  I do not understand why the answer to the following problem is $0$.
$$
\lim_{ x \to \frac{\pi}{2}+} e^{\tan x}
$$
Since $\tan(\pi/2)$ obviously does not exist, I don't understand how to determine what the limit is from the right side.  All the explanations I have found online don't really make sense so I would really appreciate an easy to understand response.  Thanks.
 A: The main point is that $x$ is tending to $\frac{\pi}{2}$ from the right, from the graph of tan $x$, when $x$ is very near to $\frac{\pi}{2}$ but greater than $\frac{\pi}{2}$, the value of tan x decreases to a number as negative as you wish, which is $-\infty$ naively, and $e^{-\infty}$=0 as you can interpret that as $e$ to a very negative number which tends to 0
A: Since $y=e^x$ is a continuous function we have:
$$
\lim_{x\to \pi/2} e^{\tan x}=e^{\lim_{x\to \pi/2} \tan x}
$$
You are correct to say that $\lim_{x\to \pi/2} \tan x$ does not exists, but there exists the two limits ( from left and from right):
$$
\lim_{x\to \pi/2^-} \tan x=+\infty \quad and \quad \lim_{x\to \pi/2^+} \tan x=-\infty
$$ 
so there exists also two limits for the exponential of $\tan x$:
$$
e^{\lim_{x\to \pi/2^-} \tan x}=e^{+\infty}=+\infty \quad and \quad e^{\lim_{x\to \pi/2^+} \tan x}=e^{-\infty}=0
$$
A: The simple explanation comes from using the basic definition of the tangent function. $$ \lim_{x \to \pi/2} \sin(x) = 1 \,\,\, \text{and}\,\,\, \lim_{x \to \pi/2} \cos(x) = 0 $$
Since you're just in calculus it should suffice to just think of $\frac{1}{0}$ as equivalent to $\pm \infty$.
Now recall that $\cos(x)$ is negative if $\pi/2<x<\pi$.
Therefore, numbers really close to $\pi/2$ and greater than $\pi/2$ will yield values for the tangent function that approach -$\infty$. As others have said already, this gives the exponential function a value of $0$.
