Limit as $\lim\limits_{x\to 0+} x^2\cot( x )$ Why does  $x^2\cot(x)$ become $0$ as $x$ tends to $0+$?
I tried using L'Hôpital's rule but I'm not getting it! Please help!! 
I'm getting the value as infinity...I think I went wrong somewhere...please help me sort it out.
 A: $$x^2\cot(x)=\frac{x^2}{\tan(x)} $$
If we directly plug in $0$ for $x$ we get $\frac{0}{0}$. So using L'Hopital's Rule:
$$\lim_{x\to 0^+} \frac{x^2}{\tan(x)} = \lim_{x \to 0^+} \frac{2x}{\sec^2(x)} = \frac{2\cdot 0}{\sec^2(0)} = \frac{0}{1} = 0$$
A: $$
x^2 \cot x = x\cos x\cdot\frac 1 {(\sin x)/x}
$$
If you can find the limits as $x\to0$ of $x$ and of $\cos x$ and of $(\sin x)/x$ then you've got it.
A: METHOD 1:
Recall that $\lim_{x\to 0} \frac{\tan x}{x}=1$.  Then, 
$$\begin{align}
\lim_{x\to 0}x^2\cot x&=\left(\lim_{x\to 0}\frac{1}{\frac{\tan x}{x}}\right)\left(\lim_{x\to 0}x\right)\\\\
&=(1)(0)\\\\
&=0
\end{align}$$

METHOD 2:
We can use asymptotics to write $\cot x=\frac1x+O(x)$.  Thus, 
$$x^2\cot x=x+O(x^3)\to 0$$

METHOD 3:
And L'Hospital's Rule gives
$$\lim_{x\to 0}x^2\cot x=\lim_{x\to 0}\frac{x^2}{\tan x}=\lim_{x\to 0}\frac{2x}{\sec^2 x}=0$$
A: In the same spirit as in answers and comments $$x^2\cot(x)=\frac{x^2}{\tan(x)}=x\frac{x}{\tan(x)}=\frac{x}{\frac{\tan(x)}x}$$ Another way using Taylor $$x^2\cot(x)=x^2\Big(\frac{1}{x}-\frac{x}{3}-\frac{x^3}{45}+O\left(x^4\right)\Big)=x-\frac{x^3}{3}-\frac{x^5}{45}+O\left(x^6\right)$$ which shows the limit and how it is approached.
