Question

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

Answer 1

$\underset{x\to 0}{lim}\frac{1-{\cos }^{3}x}{x\sin 2x}$

By L'Hospital's rule

$\underset{x\to 0}{lim}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to 0}{lim}\frac{f^{\prime }\left(x\right)}{g^{\prime }\left(x\right)}$

$\Rightarrow$ $\underset{x\to 0}{lim}\frac{0-3{\cos }^{2}x(-\sin x)}{2x\cos 2x+\sin 2x}$ [ Taking derivative of numerator and denominator ]

$\Rightarrow$ $\underset{x\to 0}{lim}\frac{3{\cos }^{2}x\sin x}{2x({\cos }^{2}x-{\sin }^{2}x)+2\sin x\cos x}$

$\Rightarrow$ $\frac{3}{2}\underset{x\to 0}{lim}\frac{{\cos }^{2}x\sin x}{x({\cos }^{2}x-{\sin }^{2}x)+2\sin x\cos x}$

$\Rightarrow$ $\frac{3}{2}\underset{x\to 0}{lim}\frac{\sin x}{x(1-{\tan }^{2}x)+2\tan x}$ [ Dividing numerator and denominator by ${\cos }^{2}x$ ]

$\Rightarrow$ $\frac{3}{2}\underset{x\to 0}{lim}\frac{\cos x}{x(-2\tan x{\sec }^{2}x)+(1-{\tan }^{2}x).1+{\sec }^{2}x}$ [ Taking derivative of numerator and denominator ]

$\Rightarrow$ $\frac{3}{2}\times \frac{1}{1+1}$

$\Rightarrow$ $\frac{3}{2}\times \frac{1}{2}$

$\Rightarrow$ $\frac{3}{4}$