Question

Evaluate:$\underset{x\to 0}{\lim }\frac{\tan x-\sin x}{x^3}$

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

Answer 1

The correct option is A

$\frac{1}{2}$

Explanation for the correct option:

Determine the value of $\underset{x\to 0}{\lim }\frac{\tan x-\sin x}{x^3}$

Consider the given Equation as,

$I=\underset{x\to 0}{\lim }\frac{\tan x-\sin x}{x^3}$

$$\begin{gathered} \Rightarrow I=\underset{x\to 0}{\lim }\frac{{\displaystyle \frac{\sin x}{\cos x}}-\sin x}{x^3}\mathbf{[}\mathbf{\because }\mathbf{tanx}\mathbf{=}\frac{\mathbf{sinx}}{\mathbf{cosx}}\mathbf{]} \\ \Rightarrow I=\underset{x\to 0}{\lim }\frac{{\displaystyle \sin x\left(1-\cos x\right)}}{\cos x.x^3} \\ \Rightarrow I=\underset{x\to 0}{\lim }\frac{\sin x}{x}\underset{x\to 0}{\lim }\frac{\left(1-\cos x\right)}{x^2}\underset{x\to 0}{\lim }\frac{{\displaystyle 1}}{\cos x} \\ \Rightarrow I=1\times \underset{x\to 0}{\lim }\frac{\left(1-\cos x\right)}{x^2}\times 1 \end{gathered}$$

Using the L' hospital rule

$$\begin{gathered} I=\underset{x\to 0}{\lim }\frac{1-\cos x}{x^2} \\ \Rightarrow I=\underset{x\to 0}{\lim }\frac{\sin x}{2x} \\ \Rightarrow I=\frac{1}{2}\mathbf{[}\mathbf{\text{Where,}}\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{\mathbf{sin}\mathbf{}\mathbf{x}}{\mathbf{x}}\mathbf{=}\mathbf{1}\mathbf{]} \end{gathered}$$

Hence, the correct answer is option A.