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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.

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    $\begingroup$ If $f$ is continuous, then $\lim\limits_{x\to c}f(g(x))=f\left(\lim\limits_{x\to c}g(x)\right)$ $\endgroup$ – user170231 Oct 23 '15 at 20:10
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    $\begingroup$ As $x$ approaches $0$ from the right, $\tan x$ becomes large negative, so $e^{\tan x}$ appproaches $0$. $\endgroup$ – André Nicolas Oct 23 '15 at 20:10
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    $\begingroup$ Either x tends to $\pi/2$ from a higher value down or else less likely it could be a print error. $\endgroup$ – Narasimham Oct 23 '15 at 20:28
  • $\begingroup$ @EmilioNovati: Your edit changed the meaning of the question; it should be $x \to (\pi/2)^+$, according to the original text. $\endgroup$ – Hans Lundmark Oct 23 '15 at 20:30
  • $\begingroup$ Sorry! I correct it. ( so my answer is redundant) :) $\endgroup$ – Emilio Novati Oct 23 '15 at 20:32
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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

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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 $\tan x$:

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

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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 $0$.

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