How to prove that $\lim\limits_{x\to\infty}e^x\text{arccot}(x)=\infty$? 
How to prove that
  $\lim\limits_{x\to\infty}e^x\text{arccot}(x)=\infty$?

I already figured that $\frac{\text{d}}{\text{dx}}[\text{arrcot}(x)]=\frac{\text{d}}{\text{dx}}\left[\arctan\left(\frac{1}{x}\right)\right]=-\frac{1}{x^2+1}$. Now I wanted to use L'Hospitals rule after doing some algebra:$$\lim\limits_{x\to\infty}e^x\text{arccot}(x)=\lim\limits_{x\to\infty}\frac{e^x}{\frac{1}{\text{arccot}(x)}}=\lim\limits_{x\to\infty}\frac{e^x}{\dfrac{1}{\left(x^2+1\right)\operatorname{arccot}^2\left(x\right)}}$$ using it twice didn't work out aswell, what am I supposed to do?
 A: $$\lim_{x\to\infty}\dfrac{\text{arccot} x}{e^{-x}}=\lim_{x\to\infty}\dfrac{-\dfrac1{1+x^2}}{-e^{-x}}=\lim_{x\to\infty}\dfrac{e^x}{1+x^2}$$
$$=\lim_{x\to\infty}\dfrac{1+x+\dfrac{x^2}2+\dfrac{x^3}{3!}+O(x^4)}{1+x^2}$$
Divide numerator & denominator by $x^2$
A: $\text{arccot}(x)=\arctan\frac{1}{x}$, for any $x\geq 1$, is bounded between $\frac{1}{2x}$ and $\frac{1}{x}$. Since $\lim_{x\to +\infty}\frac{e^x}{x}=+\infty$, you only need squeezing.
A: By standard limits we have that
$$e^x\text{arccot}(x)=e^x\text{arctan}\left(\frac1x\right)=\frac{e^x}x\frac{\text{arctan}\left(\frac1x\right)}{\frac1x} \to \infty\cdot 1 $$
A: You may set $x = \cot t$ and consider $t \to 0^+$. So,

*

*$\boxed{\lim\limits_{x\to\infty}e^x\text{arccot}(x) = \lim\limits_{t\to0^+}e^{\cot t}\cdot t}$
Furthermore you may use that for $y \geq 0$ you have


*$e^y = 1 + y +\frac{y^2}{2} + \cdots > \frac{y^2}{2}$
So for $t \rightarrow 0^+$ you get
\begin{eqnarray*} \boxed{e^{\cot t}\cdot t}
 & > & \frac{1}{2}\cot^2 t \cdot t \\
& = & \frac{\cos^2 t}{2}\frac{t^2}{\sin^2 t}\frac{1}{t} \\
& \boxed{\stackrel{t \to 0^+}{\sim}} & \boxed{\frac{1}{2t}}
\end{eqnarray*}
