Question

The value of $a$ for which $\frac{\sin 2x+a\sin x}{x^3}$ tends to a finite limit as $x\to 0$ is

• A. 1
• B. 2
• C. $-3$
• D. none of these

Answer 1

Answer: (D)

none of these

Since numerator as well as denominator both tends to zero as $x\to 0$, so applying L's Hospital's Rule, we have
$\underset{x\to 0}{lim}\frac{\sin 2x+a\sin x}{x^3}=\underset{x\to 0}{lim}\frac{2\cos 2x+a\cos x}{3x^2}$
But the numerator tends to $2+a$ as $x\to 0$. So the last limit can be finite only if $2+a=0,$ i.e., $a=-2$. In this case the last limit will be equal to
$=\underset{x\to 0}{lim}\frac{4\sin 2x+a\sin x}{6x}=-(\frac{4}{3}+\frac{a}{6})=-1$.