Question

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

• A. $2$
• B. $1$
• C. $0$
• D. $-1$

Answer 1

Answer: (B)

$1$

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

$={lim}_{x\to 0}\frac{\sin x(2a-2\cos x)}{{\sin }^{3}x/{\cos }^{3}x}$

$={lim}_{x\to 0}\frac{2{\cos }^{3}x(a-\cos x)}{{\sin }^{2}x}=\frac{2(a-1)}{0}$

but given limit exists

$\Rightarrow 2(a-1)=0\Rightarrow a=1$

we now get $\frac{0}{0}$ form

Applying L- Hospital's rule

$=2={lim}_{x\to 0}\frac{{\cos }^{3}x\left(\sin x\right)+(a-\cos x)\left(3{\cos }^{2}x\right)(-{\sin }^{2}x)}{2\sin x\cos x}$

$={lim}_{x\to 0}\frac{{\cos }^{2}x-(a-\cos x)\left(3\cos x\right)}{1}(a=1)$

$=1-(1-1)\left(3\right)=1-0=1$