Question

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

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

Answer 1

Answer: (A)

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