Question

If the value of $\underset{x\to 0}{lim}\frac{{\tan }^{3}x-{\sin }^{3}x}{x^5}$ is $\alpha$, then $2\alpha =$

Answer 1

Taking ${\tan }^{3}x$ common, we can write the given limit as:
$\underset{x\to 0}{lim}\frac{{\tan }^{3}x}{x^3}\frac{(1-{\cos }^{3}x)}{x^2}=\underset{x\to 0}{lim}\frac{{\tan }^{3}x}{x^3}\cdot \frac{(1-\cos x)}{x^2}\cdot (1+{\cos }^{2}x+\cos x)=1\cdot \frac{1}{2}\cdot 3=\frac{3}{2}\therefore 2\alpha =3$