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?
Best Answer
$$\lim_{x\to\infty}\dfrac{\text{arccot} x}{e^{-x}}=\lim_{x\to\infty}\dfrac{-\dfrac1{1+x^2}}{-e^{-x}}=\lim_{x\to\infty}\dfrac{e^x}{1+x^2}$$
$$=\lim_{x\to\infty}\dfrac{1+x+\dfrac{x^2}2+\dfrac{x^3}{3!}+O(x^4)}{1+x^2}$$
Divide numerator & denominator by $x^2$