Question

$\underset{x\to 0}{\lim }\frac{1-\cos 2x+{\tan }^{2}x}{x\sin x}$

Answer 1

$$\begin{gathered} \underset{x\to 0}{\lim }\left[\frac{1-\cos 2x+{\tan }^{2}x}{x\sin x}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{2{\sin }^{2}x+{\tan }^{2}x}{x\sin x}\right]\left[\because 1-\cos 2\mathrm{A}=2{\sin }^2\mathrm{A}\right] \\ \mathrm{Dividing}\mathrm{numerator}\&\mathrm{denominator}\mathrm{by}{x}^2: \\ \underset{x\to 0}{\lim }\left[\frac{{\displaystyle \frac{2{\sin }^{2}x}{x^2}}+{\displaystyle \frac{{\tan }^{2}x}{x^2}}}{\left({\displaystyle \frac{\sin x}{x}}\right)}\right] \\ =\frac{2{\left(1\right)}^2+{\left(1\right)}^2}{1}\left[\because \underset{x\to 0}{\lim }\frac{{\sin }^{2}x}{x^2}=1,\underset{x\to 0}{\lim }\frac{{\tan }^{2}x}{x^2}=1\right] \\ =3 \end{gathered}$$