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)

$\underset{x\to 0}{\lim }\frac{2a\sin x-\sin 2x}{{\tan }^{3}x}$
$=\underset{x\to 0}{\lim }\frac{2a\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots \right)-\left(2x-\frac{(2x{)}^3}{3!}+\frac{(2x{)}^5}{5!}-\dots \right)}{{\left(x+\frac{x^3}{3}+\frac{2}{15}{x}^5+\dots \right)}^3}$
$=\underset{x\to 0}{\lim }\frac{(2a-2)x+\left(-\frac{2a}{3!}+\frac{2^3}{3!}\right)x^3+\left(\frac{2a}{5!}-\frac{2^5}{5!}\right)x^5+\dots }{x^3{\left(1+\frac{x^2}{3}+\frac{2}{15}{x}^4+\dots \right)}^3}$
Since, it is given that given limit, is exist, then
$2a-2=0$
$\Rightarrow \alpha =1$