Question

If for some real number $a$, $\underset{x\to 0}{\text{lim}}\frac{\text{sin}2x+a\text{sin}x}{x^3}$ exists then the limit is equal to

• A. $2$
• B. $1$
• C. $-1$
• D. $-2$

Answer 1

The correct option is C

$-1$

$\underset{x\to 0}{\text{lim}}\frac{\text{sin}2x+a\text{sin}x}{x^3}=\underset{x\to 0}{\text{lim}}\frac{2\text{cos}2x+a\text{cos}x}{3x^2}$ [L'Hospital's Rule]

To make it $\left(\frac{0}{0}\right)$ form, $2\cos 2x+a\cos x$ must be equal to zero at $x=0$

$\Rightarrow 2+a=0\Rightarrow a=-2$

$\Rightarrow \underset{x\to 0}{\text{lim}}\frac{-4\text{sin}2x+2\text{sin}x}{6x}=\underset{x\to 0}{\text{lim}}[-\frac{4}{3}\frac{\text{sin}2x}{2x}+\frac{1}{3}\frac{\text{sin}x}{x}]$

$=-\frac{4}{3}+\frac{1}{3}=-1$