${\lim }_{x\to 0}\frac{1-\cos (2x)}{x\tan (x)}$

Solution

1. We are given the limit ${\lim }_{x\to 0}\frac{1-\cos (2x)}{x\tan (x)}$.
2. Using the double-angle identity $1-\cos (2x)=2{\sin }^2(x)$, the expression becomes ${\lim }_{x\to 0}\frac{2{\sin }^2(x)}{x\tan (x)}$.
3. We rewrite the limit as $2\cdot {\lim }_{x\to 0}\left(\frac{\sin (x)}{x}\right)\cdot \left(\frac{\sin (x)}{\tan (x)}\right)$.
4. Since ${\lim }_{x\to 0}\frac{\sin (x)}{x}=1$ and $$\begin{array}{rl}\underset{x\to 0}{\lim }\frac{\sin (x)}{\tan (x)}& =\underset{x\to 0}{\lim }\cos (x)\\ & =1\end{array}$$, the expression evaluates to $2(1)(1) = 2$.
5. Hence the answer is (B).