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How can I determine the limit as $x \rightarrow \infty$ of:

$\tan^{-1} * e^x$

First of all, I can't conceptualize this...I didn't think a trigonometric function could have a limit because it is constantly changing.

I also just don't even know which steps to take. $e^\infty$ will be infinity, but what is $\frac\cos\sin$ of infinity?

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  • $\begingroup$ What do you mean by $\tan^{-1}*e^x$? $\endgroup$ – MRobinson Oct 10 '18 at 15:56
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    $\begingroup$ $\tan^{-1}$ should be understood as $\arctan$ $\endgroup$ – Andrei Oct 10 '18 at 16:02
  • $\begingroup$ Also, There is no $*$ $\endgroup$ – Andrei Oct 10 '18 at 16:03
  • $\begingroup$ * is commonly used to denote multiplication in many programming languages and such. He is just new, and doesn't know how to use LaTeX to format math. Also, $\frac{1}{\tan x}$ is denoted as $\cot x$ to distinguish between that and the inverse tangent function. $\endgroup$ – HackerBoss Oct 10 '18 at 16:04
  • $\begingroup$ But then $\cot$ of what? $\endgroup$ – Andrei Oct 10 '18 at 16:06
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I will assume you want to find $$\lim_{x\rightarrow\infty}e^x\tan^{-1} x$$ (since the question was not clearly stated, and I cannot edit since another edit is pending). $\tan^{-1} x$ will go toward $\frac{\pi}{2}$ (inverse trig is different than trig, and $\tan^{-1} x\not=\frac{1}{\tan x}$). $e^x$ goes toward $\infty$, so the limit is $\infty$.

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If $\tan^{-1} x$ is $\arctan x$, then $e^\infty=\infty$ and $\arctan\infty=\pi/2$, so $$\lim_{x\to\infty}\tan^{-1}e^x=\frac{\pi}{2}$$

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