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Mathematics Question on limits and derivatives

lim⁡x→0tan⁡x−sin⁡xx3=\lim_{x\to0} \frac{\tan x -\sin x}{x^{3}} =

A

00

B

11

C

−12 - \frac{1}{2}

D

12\frac{1}{2}

Answer

12\frac{1}{2}

Explanation

Solution

lim⁡x→0tan⁡x−sin⁡xx3(00from)\displaystyle\lim_{x\to0} \frac{\tan x -\sin x}{x^{3}} \left(\frac{0}{0} from\right)
Applying L-Hospital's rule
lim⁡x→0sec⁡2x−cos⁡x3x2(00from)\displaystyle\lim _{x\to 0} \frac{\sec^{2} x -\cos x}{3x^{2}} \left(\frac{0}{0} from\right)
Again applying L-Hospital's rule,
=lim⁡x→02sec⁡xsec⁡xtan⁡x+sin⁡x6x= \displaystyle\lim _{x\to 0} \frac{2 \sec x \sec x \tan x + \sin x}{6x}
=lim⁡x→02sec⁡2xlim⁡x→0tan⁡xx+lim⁡x→0sin⁡xx6=\frac{ \displaystyle\lim _{x\to 0} 2 \sec ^{2} x \displaystyle\lim _{x\to 0} \frac{\tan x}{x} + \displaystyle\lim _{x\to 0} \frac{\sin x}{x}}{6}
=2.1.1+16=36=12= \frac{2.1.1+1}{6} = \frac{3}{6} =\frac{1}{2}