Question

$limi{t}_{x\to \frac{\mathrm{\pi }}{2}}\frac{\left(1-\tan \left({\displaystyle \frac{x}{2}}\right)\right)\left(1-\sin \left(x\right)\right)}{\left(1+\tan \left(\frac{x}{2}\right)\right){\left(\mathrm{\pi }-2\mathrm{x}\right)}^3}=$

• A. $\frac{1}{8}$
• B. $0$
• C. $\frac{1}{32}$
• D. $\infty$

Answer 1

The correct option is C

$\frac{1}{32}$

Explanation for Correct Answer:

Finding the value:

$limi{t}_{x\to \frac{\mathrm{\pi }}{2}}\frac{\left(1-\tan \left({\displaystyle \frac{x}{2}}\right)\right)\left(1-\sin \left(x\right)\right)}{\left(1+\tan \left(\frac{x}{2}\right)\right){\left(\mathrm{\pi }-2\mathrm{x}\right)}^3}=$

When we substitute limit it becomes the value $\frac{0}{0}$ indeterminate

$$\begin{gathered} limi{t}_{x\to \frac{\mathrm{\pi }}{2}}\frac{\left(1-\tan \left({\displaystyle \frac{x}{2}}\right)\right)\left(1-\sin \left(x\right)\right)}{\left(1+\tan \left(\frac{x}{2}\right)\right){\left(\mathrm{\pi }-2\mathrm{x}\right)}^3}=limi{t}_{x\to \frac{\mathrm{\pi }}{2}}\left(\frac{1-\tan \left(\frac{x}{2}\right)}{1+\left(1\right).\tan \left(\frac{x}{2}\right)}\right)\left(\frac{\left(1-\sin \left(x\right)\right)}{{\left(\mathrm{\pi }-2\mathrm{x}\right)}^3}\right) \\ =limi{t}_{x\to \frac{\mathrm{\pi }}{2}}\left(\frac{\tan \left({\displaystyle \frac{\mathrm{\pi }}{4}}\right)-\tan \left(\frac{x}{2}\right)}{1+\tan \left(\frac{\mathrm{\pi }}{4}\right)\tan \left(\frac{x}{2}\right)}\right)\left(\frac{\left(1-\sin \left(x\right)\right)}{{\left(\mathrm{\pi }-2\mathrm{x}\right)}^3}\right)\left[\because \tan \left(\frac{\mathrm{\pi }}{4}\right)=1\right] \end{gathered}$$

We know that $\tan \left(A-B\right)=\frac{\tan \left(A\right)-\tan \left(B\right)}{1+\tan \left(A\right)\tan \left(B\right)}$

Therefore,

$$\begin{gathered} =limi{t}_{x\to \frac{\mathrm{\pi }}{2}}\left(\frac{\tan \left(\frac{\mathrm{\pi }}{4}-\frac{x}{2}\right)\left(1-\sin \left(x\right)\right)}{{\left(\mathrm{\pi }-2\mathrm{x}\right)}^3}\right) \\ =limi{t}_{x\to \frac{\mathrm{\pi }}{2}}\left(\frac{\tan \left(\frac{\mathrm{\pi }-2\mathrm{x}}{4}\right)\left(1-\sin \left(x\right)\right)}{\left(\mathrm{\pi }-2\mathrm{x}\right){\left(\mathrm{\pi }-2\mathrm{x}\right)}^2}\right) \end{gathered}$$

Multiplying and dividing by $4$ in the denominator

$$\begin{gathered} =limi{t}_{x\to \frac{\mathrm{\pi }}{2}}\left(\frac{\tan \left(\frac{\mathrm{\pi }-2\mathrm{x}}{4}\right)\left(1-\sin \left(x\right)\right)}{{\displaystyle \frac{4\left(\mathrm{\pi }-2\mathrm{x}\right)}{4}}{\left(\mathrm{\pi }-2\mathrm{x}\right)}^2}\right) \\ =limi{t}_{x\to \frac{\mathrm{\pi }}{2}}\frac{1}{4}\left(\frac{\tan \left(\frac{\mathrm{\pi }-2\mathrm{x}}{4}\right)}{\frac{\left(\mathrm{\pi }-2\mathrm{x}\right)}{4}}\right)\left(\frac{\left(1-\sin \left(x\right)\right)}{{\displaystyle {\left(\mathrm{\pi }-2\mathrm{x}\right)}^2}}\right) \\ =\frac{1}{4}limi{t}_{x\to \frac{\mathrm{\pi }}{2}}\left(\frac{\tan \left(\frac{\mathrm{\pi }-2\mathrm{x}}{4}\right)}{\frac{\left(\mathrm{\pi }-2\mathrm{x}\right)}{4}}\right).limi{t}_{x\to \frac{\mathrm{\pi }}{2}}\left(\frac{\left(1-\sin \left(x\right)\right)}{{\displaystyle {\left(\mathrm{\pi }-2\mathrm{x}\right)}^2}}\right) \\ =\frac{1}{4}{\text{I}}_1.{\text{I}}_2\to \left(i\right) \end{gathered}$$

${\text{I}}_1=limi{t}_{x\to \frac{\mathrm{\pi }}{2}}\left(\frac{\tan \left(\frac{\mathrm{\pi }-2\mathrm{x}}{4}\right)}{\frac{\left(\mathrm{\pi }-2\mathrm{x}\right)}{4}}\right)$

Let's choose $y=\frac{\left(\mathrm{\pi }-2\mathrm{x}\right)}{4}$

Limit $x\to \frac{\mathrm{\pi }}{2}$, $\Rightarrow y\to 0\left[\because y=\frac{\left(\mathrm{\pi }-2{\displaystyle \frac{\mathrm{\pi }}{2}}\right)}{4}=0\right]$

$$\begin{gathered} {\text{I}}_1=limi{t}_{y\to 0}\frac{\tan \left(y\right)}{y} \\ =1\left[limi{t}_{y\to 0}\frac{\tan \left(y\right)}{y}=1\right] \end{gathered}$$

${\text{I}}_2=limi{t}_{x\to \frac{\mathrm{\pi }}{2}}\left(\frac{\left(1-\sin \left(x\right)\right)}{{\displaystyle {\left(\mathrm{\pi }-2\mathrm{x}\right)}^2}}\right)$

Using L hospitals rule

$$\begin{gathered} {\text{I}}_2=limi{t}_{x\to \frac{\mathrm{\pi }}{2}}\left(\frac{\left(1-\sin \left(x\right)\right)}{{\displaystyle {\left(\mathrm{\pi }-2\mathrm{x}\right)}^2}}\right) \\ =limi{t}_{x\to \frac{\mathrm{\pi }}{2}}\left(\frac{0-\cos \left(x\right)}{{\displaystyle 2\left(\mathrm{\pi }-2\mathrm{x}\right).(-2)}}\right)\left[\because \frac{d}{dx}\sin \left(x\right)=\cos \left(x\right);\frac{d}{dx}{x}^n=n{x}^{n-1}\right] \\ =\frac{1}{4}limi{t}_{x\to \frac{\mathrm{\pi }}{2}}\left(\frac{\cos \left(x\right)}{{\displaystyle \left(\mathrm{\pi }-2\mathrm{x}\right)}}\right) \end{gathered}$$

When we substitute limit it becomes the value $\frac{0}{0}$ indeterminate, so again applying the L hospitals rule

$$\begin{gathered} =\frac{1}{4}limi{t}_{x\to \frac{\mathrm{\pi }}{2}}\left(\frac{\cos \left(x\right)}{{\displaystyle \left(\mathrm{\pi }-2\mathrm{x}\right)}}\right) \\ =\frac{1}{4}limi{t}_{x\to \frac{\mathrm{\pi }}{2}}\frac{-\sin \left(x\right)}{{\displaystyle -2}}\left[\because \frac{d}{dx}\cos \left(x\right)=-\sin \left(x\right)\right] \\ =\frac{1}{4}.\frac{1}{{\displaystyle 2}}.\sin \left(\frac{\mathrm{\pi }}{2}\right) \\ =\frac{1}{8}\left[\because \sin \left(\frac{\mathrm{\pi }}{2}\right)=1\right] \end{gathered}$$

Substituting the value of ${\text{I}}_1$ and ${\text{I}}_2$ in equation $\left(i\right)$

$$\begin{gathered} limi{t}_{x\to \frac{\mathrm{\pi }}{2}}\frac{\left(1-\tan \left({\displaystyle \frac{x}{2}}\right)\right)\left(1-\sin \left(x\right)\right)}{\left(1+\tan \left(\frac{x}{2}\right)\right){\left(\mathrm{\pi }-2\mathrm{x}\right)}^3}=\frac{1}{4}{\text{I}}_1.\text{I} \\ =\frac{1}{4}.1.\frac{1}{8} \\ =\frac{1}{32} \end{gathered}$$

Therefore, $limi{t}_{x\to \frac{\mathrm{\pi }}{2}}\frac{\left(1-\tan \left({\displaystyle \frac{x}{2}}\right)\right)\left(1-\sin \left(x\right)\right)}{\left(1+\tan \left(\frac{x}{2}\right)\right){\left(\mathrm{\pi }-2\mathrm{x}\right)}^3}=$$\frac{1}{32}$

Hence, the correct answer is option (C).