Question

If $f\left( x \right)=\dfrac{\cos x}{{{\left( 1-\sin x \right)}^{\dfrac{1}{3}}}}$, then
A. $\underset{x\to \dfrac{\pi }{2}}{\mathop{\lim }}\,f\left( x \right)=-\infty$
B. $\underset{x\to \dfrac{\pi }{2}}{\mathop{\lim }}\,f\left( x \right)=1$
C. $\underset{x\to \dfrac{\pi }{2}}{\mathop{\lim }}\,f\left( x \right)=\infty$
D. None of these

Answer 1

Hint : We first change the variable with the relation $x=h+\dfrac{\pi }{2}$. We also change the limit and find the trigonometric values. Then we use the submultiple formulas like $\left( 1-\cosh \right)=2{{\sin }^{2}}\dfrac{h}{2}$ and $\sin \left( h \right)=2\sin \dfrac{h}{2}\cos \dfrac{h}{2}$. We find the limit value placing the $h=0$.

Complete step-by-step answer :
We first change the variable for the limit $\underset{x\to \dfrac{\pi }{2}}{\mathop{\lim }}\,f\left( x \right)$ with $f\left( x \right)=\dfrac{\cos x}{{{\left( 1-\sin x \right)}^{\dfrac{1}{3}}}}$.
As $x\to \dfrac{\pi }{2}$, we take $h=x-\dfrac{\pi }{2}\to 0$. We get $x=h+\dfrac{\pi }{2}$
So, $\underset{x\to \dfrac{\pi }{2}}{\mathop{\lim }}\,\dfrac{\cos x}{{{\left( 1-\sin x \right)}^{\dfrac{1}{3}}}}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\cos \left( h+\dfrac{\pi }{2} \right)}{{{\left( 1-\sin \left( h+\dfrac{\pi }{2} \right) \right)}^{\dfrac{1}{3}}}}$.
We have $\cos \left( h+\dfrac{\pi }{2} \right)=-\sin \left( h \right);\sin \left( h+\dfrac{\pi }{2} \right)=\cos \left( h \right)$.
So, $\underset{x\to \dfrac{\pi }{2}}{\mathop{\lim }}\,f\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{-\sin \left( h \right)}{{{\left( 1-\cosh \right)}^{\dfrac{1}{3}}}}$.
We know that $\left( 1-\cosh \right)=2{{\sin }^{2}}\dfrac{h}{2}$ and $\sin \left( h \right)=2\sin \dfrac{h}{2}\cos \dfrac{h}{2}$.

& \underset{h\to 0}{\mathop{\lim }}\,\dfrac{-\sin \left( h \right)}{{{\left( 1-\cosh \right)}^{\dfrac{1}{3}}}} \\\ & =\underset{h\to 0}{\mathop{\lim }}\,\dfrac{2\sin \dfrac{h}{2}\cos \dfrac{h}{2}}{{{\left( 2{{\sin }^{2}}\dfrac{h}{2} \right)}^{\dfrac{1}{3}}}} \\\ & =\underset{h\to 0}{\mathop{\lim }}\,\left( {{2}^{\dfrac{2}{3}}}{{\left( \sin \dfrac{h}{2} \right)}^{\dfrac{1}{3}}}\cos \dfrac{h}{2} \right) \\\ \end{aligned}$$ We place the limits and get $$\underset{x\to \dfrac{\pi }{2}}{\mathop{\lim }}\,f\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( {{2}^{\dfrac{2}{3}}}{{\left( \sin \dfrac{h}{2} \right)}^{\dfrac{1}{3}}}\cos \dfrac{h}{2} \right)=0$$. This is because as h tends to zero sine value becomes zero and we know that anything multiplied by zero will be zero. **So, the correct answer is “Option D”.** **Note** : We could also have used the main formulas for the main expression where $$\cos x={{\cos }^{2}}\dfrac{x}{2}-{{\sin }^{2}}\dfrac{x}{2}$$ and $$1-\sin x={{\left( \cos \dfrac{x}{2}-\sin \dfrac{x}{2} \right)}^{2}}$$. Then we simplify the expression to place the limit value and find its solution. In mathematics, a limit is the value that a function approaches as the input approaches some value. Limits are essential to calculus and mathematical analysis, and are used to define continuity, derivatives, and integrals.