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

Hint: 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,
\[\dfrac{1}{
  \sqrt {5 - 4\cos \left( 0 \right)} + 1 \\
   \Rightarrow \dfrac{1}{{\sqrt 1 + 1}} \\
 }\]
\[ = \dfrac{1}{2}\]
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.