Question

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

Answer 1

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

$=\underset{x\to 0}{lt}\frac{2{\sin }^{2}x+{\tan }^{2}x}{x\sin x}$

$=\underset{x\to 0}{lt}\frac{{\sin }^{2}x(2+\frac{1}{{\cos }^{2}x})}{x\sin x}$

$=\underset{x\to 0}{lt}\frac{\sin x}{x}\cdot \underset{x\to 0}{lt}(2+{\sec }^{2}x)$

$=1(2+{\sec }^{2}0)=2+1=3$.