Question

$\underset{x\to 0}{lim}\frac{{\tan }^{4}x-{\sin }^{4}x}{x^4}=$

• A. $0$
• B. $\frac{1}{6}$
• C. $\frac{2}{3}$
• D. $\frac{3}{2}$

Answer 1

The correct option is A $0$

$\underset{x\to 0}{lt}\frac{{\tan }^{4}x-{\sin }^{4}x}{x^4}$

by L H rule

$\underset{x\to 0}{lt}\frac{4{\tan }^{3}x{\sec }^{2}x-4{\sin }^{3}x\cdot \cos x}{4x^3}$

$\underset{x\to 0}{lt}\frac{{\tan }^{3}x{\sec }^{2}x}{x^3}-\frac{{\sin }^{3}x\cos x}{x^3}$

$\underset{x\to 0}{lt}{\sec }^{2}x{\left(\frac{\tan x}{x}\right)}^3-{\left(\frac{\sin x}{x}\right)}^3\cos x$

as $\underset{x\to 0}{lt}\frac{\tan x}{x}=1$ $\frac{\sin x}{x}=1$

$1-1=0$.