Question

If $\underset{x\to 0}{\lim }\frac{\sin 2x+a\sin x}{x^3}$ be finite, then the value of $a$ and the limit are given by

• A. - 2 , 1
• B. - 2 , -1
• C. 2 , 1
• D. 2 , -1

Answer 1

Answer: (B)

Let $k=\underset{x\to 0}{\lim }\frac{\sin 2x+a\sin x}{x^3}$
$=\underset{x\to 0}{\lim }\frac{2\cos 2x+a\cos x}{3x^2}$
[Using L’Hospital’s Rule]
We require $2\cos 2x+a\cos x=0$ for $x=0$ as denominator is zero.
$\therefore a=-2$
Hence, $k=\underset{x\to 0}{\lim }\frac{2\cos 2x-2\cos x}{3x^2}\left(\frac{0}{0}\right)$
$=\underset{x\to 0}{\lim }\frac{-4\sin 2x+2\sin x}{6x}\left(\frac{0}{0}\right)$
$=\underset{x\to 0}{\lim }\frac{-8\cos 2x+2\cos x}{6}$
$=\frac{-8+2}{6}=-1.$