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. 
 A: Take $y = 1/x$.
$$\frac{1 - y\cot y}{y^2} = \frac{1 - \overbrace{y\left(\frac{1}{y} - \frac{y}{3} + o(y)\right)}^{\text{Taylor expansion of $\cot$ near $0$}}}{y^2} = \frac{1}{3} - o(1) \underset{y \to 0^+}{\longrightarrow} \frac{1}{3}$$ 
A: Take $u=1/x$ then $$\lim_{u\to 0}\frac{1}{u^2}-\frac{1}{u}\cot u=\lim_{u\to 0}\frac{\sin u-u\cos u}{u^2\sin u}$$ Now applying L'Hospital twice , we get $$\lim_{u\to 0}\frac{\cos u}{3\cos u-u\sin u}=\frac{1}{3}$$
A: 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
$$
x^2-x\cot\left(\frac1x\right)=\frac13+O\left(\frac1{x^2}\right)
$$
A: Hint: expand $cot (\frac {1}{x})$ using $\sin $ and $\cos $, then get rid of the $\sin $ in the denominator (try multiplying by something that equals 1). Then try letting $u=\frac {1}{x}$, change the limit and substitute  $\frac {1}{u}$ everywhere there's an $x $.
A: WithOut L'Hospital:
i will guess that the limit exist so :
$$l=\lim_{y\to 0}\frac{1-y \cot y}{y^2}$$
$$l=\lim_{y\to 0}\frac{\sin y-y \cos y}{y^2\sin y}$$
$$l=\lim_{y\to 0}\frac{\sin y-y \cos y}{y^3}$$
$y\to 2y$
$$l=\frac{1}{4}\lim_{y\to 0}\frac{\sin y \cos y-y (1-2\sin^2y)}{y^3}$$
$$l=\frac{1}{4}\lim_{y\to 0}\frac{\sin y \cos y-y+2y\sin^2y-y\cos^2y+y\cos^2y}{y^3}$$
$$l=\frac{1}{4}\lim_{y\to 0}\frac{\sin y \cos y-y\cos^2y+y\cos^2y-y+2y\sin^2y}{y^3}$$
$$l=\frac{1}{4}\lim_{y\to 0}(\frac{\cos y(\sin y -y\cos y)}{y^3}+\frac{y\cos^2y-y}{y^3}+\frac{2y\sin^2y}{y^3})$$
$$l=\frac{1}{4}(l-1+2)$$
$$l=\frac{1}{3}$$
