Limit as $\lim\limits_{x\to 0+} x^2\cot( x )$

Question

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.

Answer 1

$$x^2\cot(x)=\frac{x^2}{\tan(x)} $$ If we directly plug in $0$ for $x$ we get $\frac{0}{0}$. So using L'Hopital's Rule: $$\lim_{x\to 0^+} \frac{x^2}{\tan(x)} = \lim_{x \to 0^+} \frac{2x}{\sec^2(x)} = \frac{2\cdot 0}{\sec^2(0)} = \frac{0}{1} = 0$$

Answer 2

$$ x^2 \cot x = x\cos x\cdot\frac 1 {(\sin x)/x} $$

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.

Answer 3

METHOD 1:

Recall that $\lim_{x\to 0} \frac{\tan x}{x}=1$. Then,

$$\begin{align} \lim_{x\to 0}x^2\cot x&=\left(\lim_{x\to 0}\frac{1}{\frac{\tan x}{x}}\right)\left(\lim_{x\to 0}x\right)\\\\ &=(1)(0)\\\\ &=0 \end{align}$$

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METHOD 2:

We can use asymptotics to write $\cot x=\frac1x+O(x)$. Thus,

$$x^2\cot x=x+O(x^3)\to 0$$

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METHOD 3:

And L'Hospital's Rule gives

$$\lim_{x\to 0}x^2\cot x=\lim_{x\to 0}\frac{x^2}{\tan x}=\lim_{x\to 0}\frac{2x}{\sec^2 x}=0$$

Answer 4

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.