Question

How do you find the limit of $ \dfrac{{1 - \cos x}}{x} $ as $ x $ approaches $ 0 $

Answer 1

Hint: L’Hospital’s rule provides a technique to evaluate limits of indeterminate form. If in a given limit $ \mathop {\lim }\limits_{x \to c} \dfrac{{f(x)}}{{g(x)}} $ , substituting $ x $ with $ c $ in the function $ \dfrac{{f(x)}}{{g(x)}} $ gives an indeterminate form like $ \dfrac{0}{0} $ or $ \dfrac{\infty }{\infty } $ , we apply L’Hospital’s rule. According to this rule, we have to substitute both the functions in the numerator and denominator with their respective derivatives.
i.e., $ \mathop {\lim }\limits_{x \to c} \dfrac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to c} \dfrac{{f'(x)}}{{g'(x)}} $
And then, we again substitute $ x $ with $ c $ in the function $ \dfrac{{f'(x)}}{{g'(x)}} $ to check the value. If the value is a definite number, we have obtained the answer. But if it again yields an indeterminate form, we continue to apply the rule by substituting both the functions in the numerator and denominator with their respective derivatives till we obtain a definite value.
So check whether L’Hospital’s rule is applicable or not. If yes, try to apply and get the answer.

Complete step-by-step answer:
(i)
We are given,
 $ \mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \cos x}}{x} $
When we substitute $ x $ as $ 0 $ in $ \dfrac{{1 - \cos x}}{x} $ , we obtain:
 $ \dfrac{{1 - \cos 0}}{0} $
Since, $ \cos 0 = 1 $ , we get:
 $ \dfrac{{1 - 1}}{0} = \dfrac{0}{0} $ i.e., an indeterminate form.
(ii)
As we know that L’Hospital’s rule helps to evaluate limits of indeterminate form, we will apply it here. So according to L’Hospital’s rule,
 $ \mathop {\lim }\limits_{x \to c} \dfrac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to c} \dfrac{{f'(x)}}{{g'(x)}} $

Here, $ f(x) = 1 - \cos x $ and $ g(x) = x $
Since we know that, $ \dfrac{{d(\cos x)}}{{dx}} = - \sin x $
Therefore, $ f'(x) = 0 - ( - \sin x) $
i.e., $ f'(x) = \sin x $
and $ g'(x) = 1 $
(iii)
Now, applying L’Hospital’s rule i.e., substituting the numerator and denominator with their respective derivatives, we get:
 $ \mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \cos x}}{x} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{1} $
(iv)
Now, we have:
 $ \mathop {\lim }\limits_{x \to 0} (\sin x) $
Putting $ x = 0 $ in $ \sin x $ gives $ 0 $ which is a definite value.
Therefore, $ \mathop {\lim }\limits_{x \to 0} (\sin x) = 0 $
And hence, $ \mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \cos x}}{x} = 0 $
So, the correct answer is “0”.

Note: We have to repeat the L’Hospital’s rule till we get a definite value instead of an indeterminate form. In this case, we got it in one step, sometimes it takes a few more to obtain the answer. We could also multiply and divide the question with the conjugate of $ 1 - \cos x $ to get $ {\sin ^2}x $ in the numerator through which $ \dfrac{{\sin x}}{x} $ could be extracted by splitting up the problem and then the values of known limits could have been substituted to obtain the answer.