Question

The value of $\mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{\left[ {1 - \tan \left( {\dfrac{x}{2}} \right)} \right]\left( {1 - \sin x} \right)}}{{\left[ {1 + \tan \left( {\dfrac{x}{2}} \right)} \right]{{\left( {\pi - 2x} \right)}^3}}}$ is
A.$\infty$
B.$\dfrac{1}{8}$
C.$0$
D.$\dfrac{1}{{32}}$

Answer 1

Hint : In order to solve the limit equation, initiate with splitting the parts and replacing the value of 1 in terms of tan to solve the first operand, then using the best suitable methods and formulas simplify it further to reduce the complexity and apply the limits. Use of trigonometric and limit formulas is mandatory.
Formula used:
$\dfrac{{\left[ {\tan \left( a \right) - \tan \left( b \right)} \right]}}{{\left[ {1 + \tan \left( a \right) \times \tan \left( b \right)} \right]}} = \tan (a - b)$
$1 - \cos x = 2{\sin ^2}\dfrac{x}{2}$
$\mathop {\lim }\limits_{x \to 0} \dfrac{{\tan \left( x \right)}}{x} = 1$
$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin \left( x \right)}}{x} = 1$

Complete step-by-step answer :
We are given with an equation $\mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{\left[ {1 - \tan \left( {\dfrac{x}{2}} \right)} \right]\left( {1 - \sin x} \right)}}{{\left[ {1 + \tan \left( {\dfrac{x}{2}} \right)} \right]{{\left( {\pi - 2x} \right)}^3}}}$.
Separating the $\tan$ part and other, and we know that the limit part is also distributed.
So, the equation can be written as:
$\mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{\left[ {1 - \tan \left( {\dfrac{x}{2}} \right)} \right]}}{{\left[ {1 + \tan \left( {\dfrac{x}{2}} \right)} \right]}}\mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{\left( {1 - \sin x} \right)}}{{{{\left( {\pi - 2x} \right)}^3}}}$ ……………..(1)
Simplifying the equation:
First Part:
From trigonometric identities, we know that $\tan \dfrac{\pi }{4} = 1$, so replacing two 1 by this in the fraction $\dfrac{{\left[ {1 - \tan \left( {\dfrac{x}{2}} \right)} \right]}}{{\left[ {1 + 1 \times \tan \left( {\dfrac{x}{2}} \right)} \right]}}$, and now it can be written as:
$\Rightarrow \dfrac{{\left[ {1 - \tan \left( {\dfrac{x}{2}} \right)} \right]}}{{\left[ {1 + 1 \times \tan \left( {\dfrac{x}{2}} \right)} \right]}} = \dfrac{{\left[ {\tan \left( {\dfrac{\pi }{4}} \right) - \tan \left( {\dfrac{x}{2}} \right)} \right]}}{{\left[ {1 + \tan \left( {\dfrac{\pi }{4}} \right) \times \tan \left( {\dfrac{x}{2}} \right)} \right]}}$
Now, from trigonometric formulas, we know that $\dfrac{{\left[ {\tan \left( a \right) - \tan \left( b \right)} \right]}}{{\left[ {1 + \tan \left( a \right) \times \tan \left( b \right)} \right]}} = \tan (a - b)$, so comparing this to the equation above, we get:
$\dfrac{{\left[ {\tan \left( {\dfrac{\pi }{4}} \right) - \tan \left( {\dfrac{x}{2}} \right)} \right]}}{{\left[ {1 + \tan \left( {\dfrac{\pi }{4}} \right) \times \tan \left( {\dfrac{x}{2}} \right)} \right]}} = \tan \left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)$
Substituting this value in Equation 1, we get:
$\mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{\left[ {1 - \tan \left( {\dfrac{x}{2}} \right)} \right]}}{{\left[ {1 + \tan \left( {\dfrac{x}{2}} \right)} \right]}}\mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{\left( {1 - \sin x} \right)}}{{{{\left( {\pi - 2x} \right)}^3}}} = \mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \tan \left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)\mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{\left( {1 - \sin x} \right)}}{{{{\left( {\pi - 2x} \right)}^3}}}$
Splitting the term ${\left( {\pi - 2x} \right)^3}$ into both the operands, and we get:
$\mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{\tan \left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)}}{{\left( {\pi - 2x} \right)}}\mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{\left( {1 - \sin x} \right)}}{{{{\left( {\pi - 2x} \right)}^2}}}$
Since, we know that $\sin x = \cos \left( {\dfrac{\pi }{2} - x} \right)$, so substituting this in the above equation:
$\mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{\tan \left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)}}{{\left( {\pi - 2x} \right)}}\mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{\left( {1 - \cos \left( {\dfrac{\pi }{2} - x} \right)} \right)}}{{{{\left( {\pi - 2x} \right)}^2}}}$ …………(2)
Now, from trigonometric identities, we know that $1 - \cos x = 2{\sin ^2}\dfrac{x}{2}$.
So, comparing $\left( {1 - \cos \left( {\dfrac{\pi }{2} - x} \right)} \right)$ with $1 - \cos x = 2{\sin ^2}\dfrac{x}{2}$, we get:

$$1 - \cos \left( {\dfrac{\pi }{2} - x} \right) = 2{\sin ^2}\left( {\left( {\dfrac{{\dfrac{\pi }{2} - x}}{2}} \right)} \right) \\\ \Rightarrow 1 - \cos \left( {\dfrac{\pi }{2} - x} \right) = 2{\sin ^2}\left( {\left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)} \right) \;$$

Substituting the upper value obtained in equation (2) , we get:
$\mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{\tan \left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)}}{{\left( {\pi - 2x} \right)}}\mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{2{{\sin }^2}\left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)}}{{{{\left( {\pi - 2x} \right)}^2}}}$
Dividing and multiplying $\left( {\pi - 2x} \right)$ by $4$ and ${\left( {\pi - 2x} \right)^2}$ by ${4^2}$.
$\mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{\tan \left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)}}{{\dfrac{4}{4}\left( {\pi - 2x} \right)}}\mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{2{{\sin }^2}\left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)}}{{\dfrac{{{4^2}}}{{{4^2}}}{{\left( {\pi - 2x} \right)}^2}}}$
Taking the denominators in the brackets:
$\mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{\tan \left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)}}{{4\left( {\dfrac{{\pi - 2x}}{4}} \right)}}\mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{2{{\sin }^2}\left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)}}{{{4^2}{{\left( {\dfrac{{\pi - 2x}}{4}} \right)}^2}}}$
On further solving, we get:
$\mathop { \Rightarrow \lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{\tan \left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)}}{{4\left( {\dfrac{\pi }{4} - \dfrac{{2x}}{4}} \right)}}\mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{2{{\sin }^2}\left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)}}{{{4^2}{{\left( {\dfrac{\pi }{4} - \dfrac{{2x}}{4}} \right)}^2}}}$
$\mathop { \Rightarrow \lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{\tan \left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)}}{{4\left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)}}\mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{2{{\sin }^2}\left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)}}{{{4^2}{{\left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)}^2}}}$ …….(3)
From formulas of limits, we know that: $\mathop {\lim }\limits_{x \to 0} \dfrac{{\tan \left( x \right)}}{x} = 1$ and same for $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin \left( x \right)}}{x} = 1$, and we can see that it is same for the two operands.
So, we get:
$\mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{\tan \left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)}}{{\left( {\dfrac{\pi }{4} - \dfrac{{2x}}{4}} \right)}} = 1$ and $\mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{{{\sin }^2}\left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)}}{{{{\left( {\dfrac{\pi }{4} - \dfrac{{2x}}{4}} \right)}^2}}} = 1$
Substituting these values in the equation 3, we get:
$\mathop { \Rightarrow \lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{\tan \left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)}}{{4\left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)}}\mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{2{{\sin }^2}\left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)}}{{{4^2}{{\left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)}^2}}}$
$\Rightarrow \dfrac{1}{4} \times 1 \times \dfrac{2}{{{4^2}}} \times 1$
Solving it further, we get:

$$\Rightarrow \dfrac{1}{4} \times \dfrac{2}{{16}} \\\ \Rightarrow \dfrac{1}{{32}} \;$$

Therefore, $\mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{\left[ {1 - \tan \left( {\dfrac{x}{2}} \right)} \right]\left( {1 - \sin x} \right)}}{{\left[ {1 + \tan \left( {\dfrac{x}{2}} \right)} \right]{{\left( {\pi - 2x} \right)}^3}}}$ is $\dfrac{1}{{32}}$ .
Hence, Option D is correct.
So, the correct answer is “Option D”.

Note : Remember to use the suitable formulas when needed, otherwise it would make the equations complex to solve.
Before solving the equation, it is recommended to use L-Hospital's rule in order to check whether the equation is in indeterminate form or not, then start simplifying the formula.
It’s important to solve step by step rather than solving at once to avoid mistakes.