Question

If $\mathop {\lim }\limits_{x \to \dfrac{x}{2}} \dfrac{{\cot x - \cos x}}{{{{(\pi - 2x)}^3}}}$ is equal to
A. $\dfrac{1}{{24}}$
B. $\dfrac{1}{{16}}$
C. $\dfrac{1}{{81}}$
D. $\dfrac{1}{4}$

Answer 1

Hint : Here, we will suppose the given angle and substitute. Replace the basic trigonometric values wherever applicable. Change accordingly the given limit with respect to the assumed substitution. Use All STC rules.

Complete step-by-step answer :
Take the given function -
$\mathop {\lim }\limits_{x \to \dfrac{x}{2}} \dfrac{{\cot x - \cos x}}{{{{(\pi - 2x)}^3}}}$
Let us consider, $x = \dfrac{\pi }{2} - h$
Here, the limit also changes to zero, where “h” tends to zero.
The given equation becomes –
$\mathop {\lim }\limits_{h \to 0} \dfrac{{\cot (\dfrac{\pi }{2} - h) - \cos (\dfrac{\pi }{2} - h)}}{{{{(\pi - 2(\dfrac{\pi }{2} - h))}^3}}}$
Apply All STC rule –
Replace,

$$\tanh = \cot [(\dfrac{\pi }{2}) - h] \\\ \cos x = \cos [(\dfrac{\pi }{2}) - h] \\\$$

Like terms cancel each other from the numerator and denominator. Like terms of same value and opposite signs cancel each other.
$\Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{\tanh - \sinh }}{{{{(2h)}^3}}}$
Convert “tangent” into sine by cosine.
$\Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{\dfrac{{\sinh }}{{\cosh }} - \sinh }}{{8{h^3}}}$
Take “sinh” common from the numerator –
$\Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{\sinh \left( {\dfrac{1}{{\cosh }} - 1} \right)}}{{8{h^3}}}$
Take L.C.M. (least common multiple) in the above equation –
$\Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{\sinh \left( {\dfrac{{1 - \cosh }}{{\cosh }}} \right)}}{{8{h^3}}}$
Take “cosh” common from the denominator.
$\Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{\dfrac{{\sinh }}{{\cosh }}\left( {1 - \cosh } \right)}}{{8{h^3}}}$
Replace the formula in the above equation.
$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }},{\text{ 1 - cos}}\theta {\text{ = 2si}}{{\text{n}}^2}\dfrac{\theta }{2}$
$\Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{\tanh \left( {2{{\sin }^2}(\dfrac{h}{2})} \right)}}{{8{h^3}}}$
Split the power of “h” in the denominator –
$\Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{\tanh }}{h}\dfrac{{\left( {2{{\sin }^2}(\dfrac{h}{2})} \right)}}{{8{h^2}}}$
Take common factors from the denominator and the numerator and simplify.
We know that, $\mathop {\lim }\limits_{h \to 0} \dfrac{{\tanh }}{h} = 1$
$\Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {{{\sin }^2}(\dfrac{h}{2})} \right)}}{{4{h^2}}}$
Multiply and divide the numerator with
$\Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {{{\sin }^2}(\dfrac{h}{2})} \right)}}{{4{{\left( {\dfrac{h}{2}} \right)}^2} \cdot 4}}$
Also, $\Rightarrow \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {{{\sin }^2}(\dfrac{h}{2})} \right)}}{{{{\left( {\dfrac{h}{2}} \right)}^2}}} = 1$ , Place value in the above equation.
$\Rightarrow \dfrac{1}{{4 \cdot 4}} = \dfrac{1}{{16}}$ is the required value.
So, the correct answer is “Option B”.

Note : Do not forget to change the limit with respect to the given limit and substitution. Remember the basic trigonometric relations and correlations along with the formulas to simplify and get the resultant value. Remember the All STC rule, it is also known as the ASTC rule in geometry. It states that all the trigonometric ratios in the first quadrant ($0^\circ \;{\text{to 90}}^\circ$ ) are positive, sine and cosec are positive in the second quadrant ( $90^\circ {\text{ to 180}}^\circ$ ), tan and cot are positive in the third quadrant ( $180^\circ \;{\text{to 270}}^\circ$ ) and sin and cosec are positive in the fourth quadrant ( $270^\circ {\text{ to 360}}^\circ$ ).