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. 

Near $x=0$,
$$
\begin{align}
\tan(x)
&=x+\frac{x^3}3+O\left(x^5\right)\\
&=x\left(1+\frac{x^2}3+O\left(x^4\right)\right)
\end{align}
$$
therefore,
$$
\begin{align}
\cot(x)
&=\frac1x\left(1-\frac{x^2}3+O\left(x^4\right)\right)\\
&=\frac1x-\frac x3+O\left(x^3\right)
\end{align}
$$
So as $x\to\infty$,
$$
\cot\left(\frac1x\right)=x-\frac1{3x}+O\left(\frac1{x^3}\right)
$$
Now we see that
$$
x^2-x\cot\left(\frac1x\right)=\frac13+O\left(\frac1{x^2}\right)
$$
