Question

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

• A. $-\frac{3}{2}$
• B. $\frac{3}{2}$
• C. $1$
• D. $0$

Answer 1

The correct option is B $\frac{3}{2}$

$\underset{x\to \pi /2}{lim}\frac{1-{\sin }^{3}x}{{\cos }^{2}x}=\underset{x\to \pi /2}{lim}\frac{(1-\sin x)(1+\sin x+{\sin }^{2}x)}{1-{\sin }^{2}x}$

$=\underset{x\to \pi /2}{lim}\frac{(1-\sin x)(1+\sin x+{\sin }^{2}x)}{(1-\sin x)(1+\sin x)}$

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