Question

$\underset{x\to 0}{\lim }\frac{\tan x-\sin x}{x^2}$ is equal to

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

Answer 1

Answer: (A)

$\underset{x\to 0}{\lim }\frac{\tan x-\sin x}{x^2},$ (form $\frac{0}{0}$)
By 'L' Hospital Rule
$\underset{x\to 0}{\lim }\frac{{\sec }^{2}x-\cos x}{2x},$ (form $\frac{0}{0}$)
Again, by 'L' Hospital Rule
$\underset{x\to 0}{\lim }\frac{2\sec x\cdot \sec x\cdot \tan x+\sin x}{2}$
$=\frac{2\cdot 1\cdot 1\cdot 0+0}{2}=\frac{0}{2}=0$