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 \[\sin \left( h \right)=2\sin \dfrac{h}{2}\cos \dfrac{h}{2}\].
\[\begin{align}
  & \underset{h\to 0}{\mathop{\lim }}\,\dfrac{-\sin \left( h \right)}{{{\left( 1-\cosh \right)}^{\dfrac{1}{3}}}} \\
 & =\underset{h\to 0}{\mathop{\lim }}\,\dfrac{2\sin \dfrac{h}{2}\cos \dfrac{h}{2}}{{{\left( 2{{\sin }^{2}}\dfrac{h}{2} \right)}^{\dfrac{1}{3}}}} \\
 & =\underset{h\to 0}{\mathop{\lim }}\,\left( {{2}^{\dfrac{2}{3}}}{{\left( \sin \dfrac{h}{2} \right)}^{\dfrac{1}{3}}}\cos \dfrac{h}{2} \right) \\
\end{align}\]
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.