Question

If $\underset{x\to 0}{lim}\frac{2a\sin x-\sin 2x}{{\tan }^{3}x}$ exists and is equal to $1$, then the value of $a$ is

• A. $2$
• B. $1$
• C. $0$
• D. $-1$

Answer 1

Answer: (B)

$1$

$\underset{x\to 0}{lim}\frac{2a\sin x-\sin 2x}{{\tan }^{3}x}$

$=\underset{x\to 0}{lim}\frac{2a(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots )-(2x-\frac{{\left(2x\right)}^3}{3!}+\frac{{\left(2x\right)}^5}{5!}-\dots )}{{(x+\frac{x^3}{3}+\frac{2}{15}{x}^5+\dots )}^3}$

$=\underset{x\to 0}{lim}\frac{(2a-2)x+(-\frac{2a}{3!}+\frac{2^3}{3!})x^3+(\frac{2a}{5!}-\frac{2^5}{5!})x^5+\dots }{x^3{(1+\frac{x^2}{3}+\frac{2}{15}{x}^4+\dots )}^3}$

Since, it is given that given limit is exist, then
$2a-2=0\Rightarrow a=1$