Question

$\underset{x\to 0}{lim}\frac{{\tan }^{4}x-{\sin }^{4}x}{x^6}=$

• A. $2$
• B. $\frac{1}{6}$
• C. $\frac{2}{3}$
• D. $\frac{3}{2}$

Answer 1

Answer: (A)

$2$

$\underset{x\to 0}{lim}\frac{{\tan }^{4}x-{\sin }^{4}x}{x^6}$
$=\underset{x\to 0}{lim}\frac{\frac{{\sin }^{4}x}{{\cos }^{4}x}-{\sin }^{4}x}{x^6}$
$=\underset{x\to 0}{lim}\frac{{\sin }^{4}x(1-{\cos }^{4}x)}{x^6{\cos }^{4}x}$
$\because {lim}_{x\to 0}\frac{1}{co{s}^{4}x}=1$
$=\underset{x\to 0}{lim}\frac{{\sin }^{4}x(1-{\cos }^{2}x)(1+{\cos }^{2}x)}{x^6}$
$=\underset{x\to 0}{lim}\frac{{\sin }^{4}x\left({\sin }^{2}x\right)(1+{\cos }^{2}x)}{x^6}$
$=\underset{x\to 0}{lim}\frac{{\sin }^{6}x(1+{\cos }^{2}x)}{x^6}=2$
Hence, option 'A' is correct.