Question & Answer
QUESTION

Evaluate the following: \[\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \sqrt {\cos x} }}{{{x^2}}}\]

ANSWER Verified Verified
Hint: Substitute the limits and check if it is of indeterminate form means when we substitute x=0 it is in $\dfrac{0}{0}$ form. If it is of indeterminate form, modify it, so we get the form \[\dfrac{{\sin x}}{x}\] , and hence, we can use \[\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1\] to evaluate.

Complete step-by-step answer:
The limit is a mathematical concept based on the idea of closeness. It is used to assign values to certain functions at points where its values are undefined, in such a way as to be consistent with nearby values.
We are asked to evaluate the function \[\dfrac{{1 - \sqrt {\cos x} }}{{{x^2}}}\] at x = 0.
We first try to directly evaluate the limit by substituting the value of x as zero.
\[\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \sqrt {\cos x} }}{{{x^2}}} = \dfrac{{1 - \sqrt {\cos 0} }}{{{0^2}}}\]
We know that the value of cos0 is equal to 1. Then, we have:
\[\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \sqrt {\cos x} }}{{{x^2}}} = \dfrac{{1 - 1}}{0}\]
\[\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \sqrt {\cos x} }}{{{x^2}}} = \dfrac{0}{0}\]
Hence, the limit is \[\dfrac{0}{0}\] form, that is, indeterminate form.
We now try to modify the function, so we can apply the limits.
We multiply the numerator and denominator by \[1 + \sqrt {\cos x} \] and then, we get using \[(a - b)(a + b) = {a^2} + {b^2}\] as follows:
\[\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \sqrt {\cos x} }}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \sqrt {\cos x} }}{{{x^2}}} \times \dfrac{{1 + \sqrt {\cos x} }}{{1 + \sqrt {\cos x} }}\]
\[\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \sqrt {\cos x} }}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{{1^2} - {{(\sqrt {\cos x} )}^2}}}{{{x^2}(1 + \sqrt {\cos x} )}}\]
\[\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \sqrt {\cos x} }}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \cos x}}{{{x^2}(1 + \sqrt {\cos x} )}}\]
We now again multiply the numerator and denominator by 1 + cosx.
\[\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \sqrt {\cos x} }}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \cos x}}{{{x^2}(1 + \sqrt {\cos x} )}} \times \dfrac{{1 + \cos x}}{{1 + \cos x}}\]
\[\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \sqrt {\cos x} }}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{1 - {{\cos }^2}x}}{{{x^2}(1 + \sqrt {\cos x} )(1 + \cos x)}}\]
We know that \[1 - {\cos ^2}x = {\sin ^2}x\], then we have:
\[\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \sqrt {\cos x} }}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{{{\sin }^2}x}}{{{x^2}(1 + \sqrt {\cos x} )(1 + \cos x)}}\]
We know that \[\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1\], using this in the above equation twice, we have:
\[\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \sqrt {\cos x} }}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} 1.1.\dfrac{1}{{(1 + \sqrt {\cos x} )(1 + \cos x)}}\]
Evaluating the limits, we have:
\[\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \sqrt {\cos x} }}{{{x^2}}} = \dfrac{1}{{(1 + \sqrt {\cos 0} )(1 + \cos 0)}}\]
\[\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \sqrt {\cos x} }}{{{x^2}}} = \dfrac{1}{{(1 + 1)(1 + 1)}}\]
\[\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \sqrt {\cos x} }}{{{x^2}}} = \dfrac{1}{{(2)(2)}}\]
\[\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \sqrt {\cos x} }}{{{x^2}}} = \dfrac{1}{4}\]
Hence, the value of the given limit is \[\dfrac{1}{4}\].

Note: For indeterminate forms, you can also evaluate the limits by using the L’Hospital rule where you can differentiate the numerator and denominator of indeterminate forms and then try to evaluate the limits.