Question

The value of $\underset{x\to \pi }{lim}\frac{1+{\cos }^{3}x}{{\sin }^{2}x}$ is

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $-\frac{1}{4}$
• D. $\frac{3}{2}$

Answer 1

Answer: (D)

$\frac{3}{2}$

$\text{We have}$

$\underset{x\to \pi }{lim}\frac{1+{\cos }^{3}x}{{\sin }^{2}x}$

$\underset{x\to \pi }{lim}\frac{1+{\cos }^{3}x}{1-{\cos }^{2}x}$
$=\underset{x\to \pi }{lim}\frac{(1+\cos x)(1-\cos x+co{s}^{2}x)}{(1-\cos x)(1+\cos x)}$ $[\because a^3+b^3=(a+b\left)\right(a^2-ab+b^2),a^2-b^2=(a+b\left)\right(a-b\left)\right]$

$=\underset{x\to \pi }{lim}\frac{1-\cos x+{\cos }^{2}x}{1-\cos x}$

$=\frac{1+1+1}{1+1}=\frac{3}{2}$