# Determining trigonometric limit

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?

• What do you mean by $\tan^{-1}*e^x$? – MRobinson Oct 10 '18 at 15:56
• $\tan^{-1}$ should be understood as $\arctan$ – Andrei Oct 10 '18 at 16:02
• Also, There is no $*$ – Andrei Oct 10 '18 at 16:03
• * 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. – HackerBoss Oct 10 '18 at 16:04
• But then $\cot$ of what? – Andrei Oct 10 '18 at 16:06

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