Question

$f\left(x\right)=\underset{n\to \infty }{lim}\frac{{\tan }^{2n}x\cdot \sin x}{x}$ then $\underset{x\to 0}{lim}f\left(x\right)=1$
If true enter 1, else enter 0.

Answer 1

$f\left(x\right)=\underset{n\to \infty }{lim}\frac{{tan}^{2n}x.\sin x}{x}g\left(x\right)=\underset{x\to 0}{lim}\underset{n\to \infty }{lim}\frac{{tan}^{2n}x.\sin x}{x}g\left(x\right)=\underset{x\to 0}{lim}{tan}^{2n}x\underset{n\to \infty }{lim}\frac{\sin x}{x}g\left(x\right)=0\times 1g\left(x\right)=0$