Question

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

• A. $0$
• B. $1$
• C. $\frac{1}{2}$
• D. none of these

Answer 1

Answer: (A)

$0$

$\underset{x\to 0}{lim}\frac{x-\tan x}{x\tan x}$

Applying L' Hospital rule, we get

$\underset{x\to 0}{lim}\frac{1-{\sec }^{2}x}{\tan x+x{\sec }^{2}x}$

Again applying L' Hospital rule, we get

$\underset{x\to 0}{lim}\frac{0-2{\sec }^{2}x\tan x}{{\sec }^{2}x+{\sec }^{2}x+2x\sec x^{2}x\tan x}=0$

Hence, option 'A' is correct.