Question

If $\underset{x\to 0}{lim}\frac{2a\sin x-\sin 2x}{{\tan }^{3}x}$ exists and is equal to $1$, then the value of $a$ is:

• A. $2$
• B. $1$
• C. $0$
• D. $-1$

Answer 1

The correct option is B

$1$

Explanation for the correct option.

Step 1: Apply limit.

We have $\underset{x\to 0}{lim}\frac{2a\sin x-\sin 2x}{{\tan }^{3}x}$. By applying limit we get,

$\frac{2a\sin 0-\sin \left(0\right)}{{\tan }^{3}0}=\frac{0}{0}$.

As it is in $\frac{0}{0}$ form, so we will apply L' Hospital's Rule.

Step 2: Apply L' Hospital's Rule.

$\begin{array}{rcl}\underset{x\to 0}{lim}\frac{2a\sin x-\sin 2x}{{\tan }^{3}x}& =& \underset{x\to 0}{lim}\frac{{\displaystyle \frac{d\left(2a\sin x\right)}{dx}}-{\displaystyle \frac{d\left(\sin 2x\right)}{dx}}}{{\displaystyle \frac{d\left({\tan }^{3}x\right)}{dx}}}\\ & =& \underset{x\to 0}{lim}\frac{{\displaystyle 2a\cos x-2\cos 2x}}{{\displaystyle 3{\tan }^{2}x·{\sec }^{2}x}}\\ & =& \frac{{\displaystyle 2a\cos 0-2\cos 0}}{{\displaystyle 3{\tan }^{2}0·{\sec }^{2}0}}\\ & =& \frac{2a-2}{0}\end{array}$

Step 3: Find the value of $a$.

Its is given that $\underset{x\to 0}{lim}\frac{2a\sin x-\sin 2x}{{\tan }^{3}x}=1$

$\begin{array}{rcl}& \Rightarrow & \frac{2a-2}{0}=1\\ \Rightarrow 2a& =& 2\\ \Rightarrow a& =& 1\end{array}$

Hence, option B is correct.