Question

How do you evaluate \[\dfrac{{1 - \cos 2x}}{{{x^2}}}\] as ‘x’ approaches 0?

Answer 1

Hint:We can solve this using two methods. First method is using L’hospital’s rule. This rule states that if we have the cases \[\mathop {\lim }\limits_{x \to a} \dfrac{{f(x)}}{{g(x)}} =
\dfrac{0}{0}{\text{ OR }}\dfrac{{ \pm \infty }}{{ \pm \infty }}\] where ‘a’ can be any real number, infinity or negative infinity. In these cases we have, \[\mathop {\lim }\limits_{x \to a}
\dfrac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to a} \dfrac{{f'(x)}}{{g'(x)}}\]. We can use this rule as many times as we need if we keep on getting indeterminate form.

Complete step by step solution:
Given, \[\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \cos 2x}}{{{x^2}}}\].
If we try to substitute 0 into \[\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \cos 2x}}{{{x^2}}}\], we will get \[\dfrac{0}{0}\]. This is an indeterminate form.
When direct substitution yields an indeterminate form, we use L’hospital’s rule:
\[\mathop {\lim }\limits_{x \to a} \dfrac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to a}
\dfrac{{f'(x)}}{{g'(x)}}\]
We differentiate the numerator term and the denominator term with respect to ‘x’.
Then applying we get,
\[ \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \cos 2x}}{{{x^2}}} = \mathop {\lim
}\limits_{x \to 0} \dfrac{{2.\sin 2x}}{{2.x}}\]
Cancelling 2 we get,
\[ = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 2x}}{x}\]
If we substitute 0 into \[\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 2x}}{x}\] we will again get \[\dfrac{0}{0}\].
This is an indeterminate form. So we again apply the L’hospital’s rule that differentiate the numerator term and the denominator term with respect to ‘x’.
\[ = \mathop {\lim }\limits_{x \to 0} \dfrac{{2.\cos 2x}}{1}\]
Taking constant outside we get,
\[ = 2.\mathop {\lim }\limits_{x \to 0} \cos 2x\]
If we substitute 0 we will get,
\[ = 2\].

Hence, \[\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \cos 2x}}{{{x^2}}} = 2\]

Note: After the substitution of limit value if we keep on getting indeterminate values we can apply as many times we need (see in above steps). In the above step we have \[ = \mathop {\lim
}\limits_{x \to 0} \dfrac{{\sin 2x}}{x}\] we can solve this using another method.
Multiplying and dividing by 2 on the denominator we get,

\[ = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 2x}}{{\left( {\dfrac{1}{2} \times 2x} \right)}}\]
Since we have the constant term \[\dfrac{1}{{\left( {\dfrac{1}{2}} \right)}} = 2\]
\[ = 2\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 2x}}{{2x}}\] (the angel term and the
denominator are same) We know that from a theorem we have \[\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} =1\]applying this we get, \[ = 2\]. The answer is the same in both cases: we can choose any method.