For calculating the limit $\lim_{x \to 0} x\cot(2x)$, why is it wrong to follow the second method?

Question

I'm preparing for my Graduate Admission Test, let's considering this limit:

$$ \lim_{x \to 0} x\cot(2x) $$

The correct answer is :

$$ \lim_{x \to 0} x\cot(2x) = \lim_{x \to 0} \frac {x}{\tan(2x)} = \lim_{x \to 0} \frac {1}{2\sec^22x} = \frac {1}{2} $$

But I'm doing like this:

$$ \lim_{x \to 0} x\cot(2x) = \lim_{x \to 0} \frac {\cot(2x)}{\frac{1}{x}} = \lim_{x \to 0} \frac {2\csc^2x}{\frac {1}{x^2}} = \lim_{x \to 0} \frac {2x^2}{\sin^2x} = 2 $$

Because

$$ \lim_{x \to 0} \frac {x}{\sin x} = 1 \Longrightarrow \lim_{x \to 0} \frac {2x^2}{\sin^2x} = 2 $$

So, what am I doing wrong? After thinking about it several times, it feels okay.

Answer 1

Thanks for @Ninad Munshi , I'm using the wrong chain rule.

Correct method is:

$$ \lim\limits_{x \to 0} x \cot\left( 2 x \right) = \lim\limits_{x \to 0} \frac{\cot\left( 2 x \right)}{\frac{1}{x}} = \lim\limits_{x \to 0} \frac{2 \csc^{2}\left( 2 x \right)}{\frac {1}{x^2}} = \lim\limits_{x \to 0} \frac{2 x^{2}}{\sin^{2}\left( 2 x \right)} = \lim\limits_{x \to 0} \frac {\left( 2 x \right)^{2}}{\sin^{2}\left( 2 x \right)} \cdot \frac{1}{2} = \frac{1}{2} $$

Answer 2

Alternatively, since $\sin(2x)\sim 2x$ when $x\to 0$, you may approach the proposed exercise as follows: \begin{align*} \lim_{x\to 0}x\cot(2x) & = \lim_{x\to 0}\frac{x\cos(2x)}{\sin(2x)}\\\\ & = \lim_{x\to 0}\left(\frac{2x}{\sin(2x)}\right)\frac{\cos(2x)}{2}\\\\ & = \lim_{x\to 0}\left(\frac{2x}{2x}\right)\frac{\cos(2x)}{2}\\\\ & = \lim_{x\to 0}\frac{\cos(2x)}{2} = \frac{1}{2} \end{align*}

Hopefully this helps!

Answer 3

By Taylor's formula, $$\frac{1}{\tan(2x)}=\frac{1}{2x+O(x^3)}$$ as $x\to0$. Multiplying by $x$ and taking the limit gives $1/2$.