Question

$\underset{x\to 0}{lim}\frac{\sin 2x+a\sin x}{x^3}=$ finite,find $a$ and the limit.

• A. $a=-2$
• B. $a=-1$
• C. limit$=-1$
• D. limit$=-2$

Answer 1

Answer: (A), (C)

The correct options are
A $a=-2$
C limit$=-1$

$L=\underset{x\to 0}{lim}\frac{\sin 2x+a\sin x}{x^3}=\underset{x\to 0}{lim}\left(\frac{\sin x}{x}\right)\left(\frac{2\cos x+a}{x^2}\right)=\underset{x\to 0}{lim}\left(\frac{2(1-\frac{x^2}{2!}+\frac{x^4}{4!}-...\infty )+a}{x^2}\right)$
L is finite when $a+2=0$
Therefore, $a=-2$
now, $L=\underset{x\to 0}{lim}\left(\frac{2(-\frac{x^2}{2!}+\frac{x^4}{4!}-...\infty )}{x^2}\right)=-1$
Ans: A,C