Question

If $f\left(x\right)=\frac{\cos x}{{(1-\sin x)}^{1/3}},$ then

• A. $\underset{x\to \frac{\pi }{2}}{lim}f\left(x\right)=1$
• B. $\underset{x\to \frac{{\pi }^{+}}{2}}{lim}f\left(x\right)$ does not exist
• C. $\underset{x\to \frac{\pi }{2}}{lim}f\left(x\right)=-1$
• D. $\underset{x\to \frac{\pi }{2}}{lim}f\left(x\right)=0$

Answer 1

The correct option is D $\underset{x\to \frac{\pi }{2}}{lim}f\left(x\right)=0$

Let $L=\underset{x\to \frac{\pi }{2}}{lim}\frac{\cos x}{{(1-\sin x)}^{1/3}}$
Put $x=\frac{\pi }{2}+t$
So, we have:
$L=-\underset{t\to 0}{lim}\frac{\sin t}{{(1-\cos t)}^{1/3}}=-\underset{t\to 0}{lim}\frac{2\sin \frac{t}{2}\cos \frac{t}{2}}{{\left(2{\sin }^2\frac{t}{2}\right)}^{\frac{1}{3}}}=-\underset{t\to 0}{lim}{2}^{2/3}\cos \frac{t}{2}{\left(\sin \frac{t}{2}\right)}^{\frac{1}{3}}=0$