Question

$\underset{x\to \frac{n}{2}}{lim}\frac{1-(\sin x{)}^{\frac{-1}{3}}}{1-(\sin x{)}^{\frac{-2}{3}}}$ is?

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

Answer 1

The correct option is C $\frac{1}{2}$

$\begin{array}{c}{lim}_{x\to \frac{\pi }{2}}\frac{1-\sin x^{-1/3}}{1-\sin x^{-2/3}}is?\\ \frac{\sin x^{1/3}-1}{\sin x^{1/3}}\times \frac{\sin x^{2/3}}{\sin x^{2/3}-1}\left[\because a^2-b^2=\left(a+b\right)\left(a-b\right)\right]\\ \left[\sin \frac{\pi }{2}=1\right]\\ \Rightarrow {lim}_{x\to \frac{\pi }{2}}\frac{\left(\sin x^{1/3}-1\right)\times \sin x^{1-3}}{{\left(\sin x^{1/3}\right)}^2-1^2}=\frac{\sin x^{1/3}\left(\sin x^{1/3}-1\right)}{\left(\sin x^{1/3}-1\right)\left(\sin x^{1/3}+1\right)}\\ {lim}_{x\to \frac{\pi }{2}}\frac{\sin x^{1/3}}{\left(\sin x^{1/3}+1\right)}=\frac{\sin x^{1/3}\frac{\pi }{2}}{\sin x^{1/3}\frac{\pi }{2}+1}=\frac{1}{1+1}=\frac{1}{2}Ans.\\ \end{array}$