Question

$\underset{x\to 0}{lim}\frac{\tan x-\sin x}{x^3}$ is equal to

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

Answer 1

The correct option is A $\frac{1}{2}$

$\underset{x\to 0}{lim}\frac{\tan x-\sin x}{x^3}\left(\frac{0}{0}form\right)$ (using L'Hospital's rule)

$=\underset{x\to 0}{lim}\frac{{\sec }^{2}x-\cos x}{3x^2}\left(\frac{0}{0}form\right)$ (using L'Hospital's rule)

$=\underset{x\to 0}{lim}\frac{2{\sec }^{2}x\tan x+\sin x}{6x}\left(\frac{0}{0}form\right)$

$=\underset{x\to 0}{lim}\frac{\left[2\right({\sec }^{2}x{\sec }^{2}x+2\sec x\times \sec x\tan x\tan x)+\cos x]}{6}$ (using L'Hospital's rule)

$=\frac{2[1.1+2(0\left)\right]+1}{6}$

$=\frac{3}{6}=\frac{1}{2}$