Question

The sum of an infinite geometric series whose first term is the limit of the function $f\left( x \right) = \dfrac{{\tan x - \sin x}}{{{{\sin }^3}x}}$ as $x \Rightarrow 0$ and whose common ratio is the limit of the function $g\left( x \right) = \dfrac{{1 - \sqrt x }}{{{{\left( {{{\cos }^{ - 1}}x} \right)}^2}}}$ as $x \Rightarrow 1$ is
A. $\dfrac{1}{3}$
B. $\dfrac{1}{4}$
C. $\dfrac{1}{2}$
D. $\dfrac{2}{3}$

Answer 1

Hint : First find the limits of the given two functions and limit of $f\left( x \right)$ is a (the first term of G.P) and limit of $g\left( x \right)$ is r (Common ratio of G.P). Sum of infinite terms of a Geometric progression is given by the below mentioned formula.
Formulas used:
1. Sum of infinite terms of a G.P is $\dfrac{a}{{1 - r}}$ where a is the first term and r is the common ratio and r must be greater than -1 and less than 1.
2. $\mathop {\lim }\limits_{x \Rightarrow 0} \dfrac{{\sin x}}{x} = 1$

Complete step-by-step answer :
We are given that the limit of the function $f\left( x \right) = \dfrac{{\tan x - \sin x}}{{{{\sin }^3}x}}$ as $x \Rightarrow 0$ is the first term and limit of the function $g\left( x \right) = \dfrac{{1 - \sqrt x }}{{{{\left( {{{\cos }^{ - 1}}x} \right)}^2}}}$ as $x \Rightarrow 1$ is the common ratio of an infinite geometric series.
We have to find the sum of this infinite geometric series.
Limit of the function $f\left( x \right) = \dfrac{{\tan x - \sin x}}{{{{\sin }^3}x}}$ as $x \Rightarrow 0$ is $\mathop {\lim }\limits_{x \Rightarrow 0} \dfrac{{\tan x - \sin x}}{{{{\sin }^3}x}}$
$\tan x$ can also be written as $\dfrac{{\sin x}}{{\cos x}}$
$\Rightarrow \mathop {\lim }\limits_{x \Rightarrow 0} \dfrac{{\left( {\dfrac{{\sin x}}{{\cos x}}} \right) - \sin x}}{{\sin x\left( {{{\sin }^2}x} \right)}}$
$\Rightarrow \mathop {\lim }\limits_{x \Rightarrow 0} \dfrac{{\left( {\dfrac{1}{{\cos x}}} \right) - 1}}{{\left( {{{\sin }^2}x} \right)}} = \Rightarrow \mathop {\lim }\limits_{x \Rightarrow 0} \dfrac{{\left( {\dfrac{{1 - \cos x}}{{\cos x}}} \right)}}{{\left( {{{\sin }^2}x} \right)}}$
$\Rightarrow \mathop {\lim }\limits_{x \Rightarrow 0} \dfrac{{1 - \cos x}}{{\cos x\left( {{{\sin }^2}x} \right)}}$
We know that ${\sin ^2}x + {\cos ^2}x = 1 \Rightarrow {\sin ^2}x = 1 - {\cos ^2}x$
$\Rightarrow \mathop {\lim }\limits_{x \Rightarrow 0} \dfrac{{1 - \cos x}}{{\cos x\left( {1 - {{\cos }^2}x} \right)}} = \mathop {\lim }\limits_{x \Rightarrow 0} \dfrac{{1 - \cos x}}{{\cos x\left( {1 - \cos x} \right)\left( {1 + \cos x} \right)}}$
$\Rightarrow \mathop {\lim }\limits_{x \Rightarrow 0} \dfrac{1}{{\cos x\left( {1 + \cos x} \right)}}$, $\cos 0 = 1$
$\Rightarrow \mathop {\lim }\limits_{x \Rightarrow 0} \dfrac{1}{{\cos x\left( {1 + \cos x} \right)}} = \dfrac{1}{{\cos 0\left( {1 + \cos 0} \right)}} = \dfrac{1}{2}$
The first term of the series is $\dfrac{1}{2}$
Limit of the function $g\left( x \right) = \dfrac{{1 - \sqrt x }}{{{{\left( {{{\cos }^{ - 1}}x} \right)}^2}}}$ as $x \Rightarrow 1$ is $\mathop {\lim }\limits_{x \Rightarrow 1} \dfrac{{1 - \sqrt x }}{{{{\left( {{{\cos }^{ - 1}}x} \right)}^2}}}$
Multiply and divide the limit with $1 + \sqrt x$
$\Rightarrow \mathop {\lim }\limits_{x \Rightarrow 1} \dfrac{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}}{{{{\left( {{{\cos }^{ - 1}}x} \right)}^2}\left( {1 + \sqrt x } \right)}} = \dfrac{1}{{\left( {1 + \sqrt 1 } \right)}}\mathop {\lim }\limits_{x \Rightarrow 1} \dfrac{{1 - x}}{{{{\left( {{{\cos }^{ - 1}}x} \right)}^2}}}$ as $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$
Let ${\cos ^{ - 1}}x = t$, this gives $x = \cos t$ and the limit changes from $x \Rightarrow 1$ to $t \Rightarrow 0$ as ${\cos ^{ - 1}}1$ is equal to zero.
Substitute the values of x and ${\cos ^{ - 1}}x$
$\Rightarrow \dfrac{1}{2}\mathop {\lim }\limits_{t \Rightarrow 0} \dfrac{{1 - \cos t}}{{{t^2}}}$
We can write $\cos t$ as $1 - 2{\sin ^2}\dfrac{t}{2}$
$\Rightarrow \dfrac{1}{2}\mathop {\lim }\limits_{t \Rightarrow 0} \dfrac{{1 - \left( {1 - 2{{\sin }^2}\left( {\dfrac{t}{2}} \right)} \right)}}{{{t^2}}} = \dfrac{{2{{\sin }^2}\left( {\dfrac{t}{2}} \right)}}{t}$
Multiplying and dividing with $\dfrac{1}{4}$
$\Rightarrow \dfrac{1}{2} \times 2 \times \dfrac{1}{4}\mathop {\lim }\limits_{t \Rightarrow 0} \dfrac{{\sin \left( {\dfrac{t}{2}} \right) \times \sin \left( {\dfrac{t}{2}} \right)}}{{\left( {\dfrac{t}{2}} \right) \times \left( {\dfrac{t}{2}} \right)}} = \dfrac{1}{4} \times 1 \times 1$ as $\mathop {\lim }\limits_{x \Rightarrow 0} \dfrac{{\sin x}}{x} = 1$ is equal to $\dfrac{1}{4}$
Therefore, the common ratio of the series is $\dfrac{1}{4}$
The sum of the infinite terms of the geometric series will be $\dfrac{a}{{1 - r}} = \dfrac{{\left( {\dfrac{1}{2}} \right)}}{{1 - \left( {\dfrac{1}{4}} \right)}} = \dfrac{{\left( {\dfrac{1}{2}} \right)}}{{\left( {\dfrac{3}{4}} \right)}} = \dfrac{1}{2} \times \dfrac{4}{3} = \dfrac{2}{3}$
Hence, the correct option is Option D, $\dfrac{2}{3}$
So, the correct answer is “Option D”.

Note : When the common ratio is greater than 1 the geometric series does not have a sum, which means the sum cannot be determined. Only when the common ratio is between -1 and 1 the above formula is applicable. Do not confuse the formulas of G.P with the formulas of A.P.