Question

Find the value of the limit of the expression $\mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {\sqrt {\left( {1 - \left. {\cos x} \right) + \sqrt {\left( {1 - \cos x} \right) + \sqrt {\left( {1 - \cos x} \right) + ...\infty } } } \right.} } \right) - 1}}{{{x^2}}}$ equals to?
A). $0$
B). $\dfrac{1}{2}$
C). $1$
D). $2$

Answer 1

First to simplify the question part to solve further, we will assume
y = $$$$\sqrt {\left( {1 - \left. {\cos x} \right) + \sqrt {\left( {1 - \cos x} \right) + \sqrt {\left( {1 - \cos x} \right) + ...\infty } } } \right.} . This will imply $\sqrt {\left( {1 - \cos x} \right) + y} = y$. Solve this equation further for $y$. Now, Put the values of $y$ in the given equation part. Further we will simplify the equation, if needed, to make the value of the limit defined i.e., make the denominator part non-zero by rationalization method.
Formula Used:
$1 - \cos \theta = 2{\sin ^2}\dfrac{\theta }{2}$
$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$
Multiplicative property of limit: $\mathop {\lim }\limits_{x \to a} f\left( x \right).g(x) = \mathop {\lim }\limits_{x \to a} f\left( x \right) \times \mathop {\lim }\limits_{x \to a} g\left( x \right)$

Complete step-by-step solution:
$\mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {\sqrt {\left( {1 - \left. {\cos x} \right) + \sqrt {\left( {1 - \cos x} \right) + \sqrt {\left( {1 - \cos x} \right) + ...\infty } } } \right.} } \right) - 1}}{{{x^2}}}$
Let us assume y = $$$$\sqrt {\left( {1 - \left. {\cos x} \right) + \sqrt {\left( {1 - \cos x} \right) + \sqrt {\left( {1 - \cos x} \right) + ...\infty } } } \right.}
$\Rightarrow y = \sqrt {\left( {1 - \cos x} \right) + y}$
Squaring on both sides, we get
${y^2} = \left( {1 - \cos x} \right) + y$
$\Rightarrow {y^2} - y + \left( {\cos x - 1} \right) = 0\;\;\;\;\;\;.......(1)$
By comparing equation (1) with the standard form of quadratic equation i.e., $a{x^2} + bx + c = 0$, we get,

$$a = 1, \\\ b = - 1, \\\ c = \cos x - 1 \\\$$

Solving this quadratic equation for the values of $y$using Discriminant,
$y = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Putting the values of $a,b$and $c$from equation (1), we get,
$y = \dfrac{{ - ( - 1) \pm \sqrt {{{( - 1)}^2} - 4(1)(\cos x - 1)} }}{{2(1)}}$
$\Rightarrow y = \dfrac{{1 \pm \sqrt {5 - 4\cos x} }}{2}$
Substitute the value of $y = \dfrac{{1 + \sqrt {5 - 4\cos x} }}{2}$ in the given function,
$\mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {\dfrac{{1 + \sqrt {5 - 4\cos x} }}{2}} \right) - 1}}{{{x^2}}}$
Simplifying it, we get

$$\mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{{1 + \sqrt {5 - 4\cos x} - 2}}{2}}}{{{x^2}}} \\\ \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {5 - 4\cos x} - 1}}{{2{x^2}}} \\\$$

Simplifying it further using rationalization,

$$\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\sqrt {5 - 4\cos x} - 1}}{{2{x^2}}} \times \dfrac{{\sqrt {5 - 4\cos x} + 1}}{{\sqrt {5 - 4\cos x} + 1}}} \right) \\\ \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{5 - 4\cos x - 1}}{{2{x^2}\left( {\sqrt {5 - 4\cos x} + 1} \right)}} \\\ \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{4\left( {1 - \cos x} \right)}}{{2{x^2}\left( {\sqrt {5 - 4\cos x} + 1} \right)}} \\\$$

Using the identity,
$\mathop {\lim }\limits_{x \to 0} \dfrac{{4\left( {2{{\sin }^2}\dfrac{x}{2}} \right)}}{{2{x^2}\left( {\sqrt {5 - 4\cos x} + 1} \right)}}$ …….$\left[ {\because 1 - \cos \theta = 2{{\sin }^2}\dfrac{\theta }{2}} \right]$
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{4{{\sin }^2}\dfrac{x}{2}}}{{{x^2}\left( {\sqrt {5 - 4\cos x} + 1} \right)}}$
Using multiplicative property of limit,
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{4{{\sin }^2}\dfrac{x}{2}}}{{{x^2}}} \times \mathop {\lim }\limits_{x \to 0} \dfrac{1}{{\left( {\sqrt {5 - 4\cos x} + 1} \right)}}$ $\left[ {\because \mathop {\lim }\limits_{x \to a} f\left( x \right).g(x) = \mathop {\lim }\limits_{x \to a} f\left( x \right) \times \mathop {\lim }\limits_{x \to a} g\left( x \right)} \right]$
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{{{\sin }^2}\dfrac{x}{2}}}{{{{\left( {\dfrac{x}{2}} \right)}^2}}} \times \mathop {\lim }\limits_{x \to 0} \dfrac{1}{{\left( {\sqrt {5 - 4\cos x} + 1} \right)}}$
Using the identity of limit,
${(1)^2} \times \mathop {\lim }\limits_{x \to 0} \dfrac{1}{{\left( {\sqrt {5 - 4\cos x} + 1} \right)}}$ $\left[ {\because \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1} \right]$
Put the final limit,

\sqrt {5 - 4\cos \left( 0 \right)} + 1 \\\ \Rightarrow \dfrac{1}{{\sqrt 1 + 1}} \\\ }$$ $$ = \dfrac{1}{2}$$ **Hence, the Final Answer: C. $$\dfrac{1}{2}$$** **Note:** If we directly put the limit in the given question, we will get the indeterminate form i.e. we get a zero in denominator. So, to solve further we have to simplify the numerator part. Also, if we take$$y = \dfrac{{1 - \sqrt {5 - 4\cos x} }}{2}$$, then again we get the indeterminate form after putting the limits to the function. So, we will not consider this value of $$y$$. Alternatively, this problem can be solved using L-Hospital’s Rule, which states: Suppose, we have one of the following cases: $$\mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \dfrac{0}{0}$$or $$\mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \dfrac{{ \pm \infty }}{{ \pm \infty }}$$, where $$a$$can be any real number or even $$ \pm \infty $$. In these cases, $$\mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to a} \dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}$$, where $$f'\left( x \right)\& g'\left( x \right)$$represents the derivatives of $$f(x)\& g(x)$$respectively.