Question

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