How to find the limit of trig functions in exponents?

Question

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.

Answer 1

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

Answer 2

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 $\tan x$:

$$ e^{\lim_{x\to \pi/2^-} \tan x}=e^{+\infty}=+\infty \quad and \quad e^{\lim_{x\to \pi/2^+} \tan x}=e^{-\infty}=0 $$

Answer 3

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 $0$.