Limit of function $x^2-x\cot\left(\frac{1}{x}\right)$ How To compute $\lim_{x \to \infty}  x^2 -x \cot(1/x)$? Wolfram says it is $\frac{1}{3}$ and I know it is supposed to converge to a number other than 0 but I keep getting infinity. 

WithOut L'Hospital:
i will guess that the limit exist so :
$$l=\lim_{y\to 0}\frac{1-y \cot y}{y^2}$$
$$l=\lim_{y\to 0}\frac{\sin y-y \cos y}{y^2\sin y}$$
$$l=\lim_{y\to 0}\frac{\sin y-y \cos y}{y^3}$$
$y\to 2y$
$$l=\frac{1}{4}\lim_{y\to 0}\frac{\sin y \cos y-y (1-2\sin^2y)}{y^3}$$
$$l=\frac{1}{4}\lim_{y\to 0}\frac{\sin y \cos y-y+2y\sin^2y-y\cos^2y+y\cos^2y}{y^3}$$
$$l=\frac{1}{4}\lim_{y\to 0}\frac{\sin y \cos y-y\cos^2y+y\cos^2y-y+2y\sin^2y}{y^3}$$
$$l=\frac{1}{4}\lim_{y\to 0}(\frac{\cos y(\sin y -y\cos y)}{y^3}+\frac{y\cos^2y-y}{y^3}+\frac{2y\sin^2y}{y^3})$$
$$l=\frac{1}{4}(l-1+2)$$
$$l=\frac{1}{3}$$
