Question

$L{t}_{x\to 0}\frac{\sin 2x+a\sin x}{x^3}$ exists and finite then $a=$

• A. $2$
• B. $-2$
• C. $\frac{2}{3}$
• D. $\frac{-2}{3}$

Answer 1

The correct option is B $-2$

$L{t}_{x\to 0}\frac{\sin 2x+a\sin x}{x^3}$

$L{t}_{x\to 0}\frac{\sin x(2\cos x+a)}{x^3}$

$=L{t}_{x\to 0}\frac{(2\cos x+a)}{x^2}$

As $x\to 0$, denominator tends to 0, so the numerator also tends to 0.

$\Rightarrow L{t}_{x\to 0}2\cos x+a=0$

$\Rightarrow a=-2$

Hence, option 'B' is correct.