Question

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

• A. $2$
• B. $1$
• C. $0$
• D. $-1$

Answer 1

Answer: (B)

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