Question

How do you evaluate the limit $\dfrac{{1 - \cos x}}{x}$ as x approaches zero ?

Answer 1

Hint : The given question requires us to evaluate a limit that is of indeterminate form. Limits of indeterminate form can be easily solved using the L’Hospital’s rule. There are various methods and steps to evaluate a limit. Some of the common steps while solving limits involve rationalization and applying some basic results on frequently used limits. L’Hospital’s rule involves solving limits of indeterminate form by differentiating both numerator and denominator with respect to the variable separately and then applying the required limit.

Complete step-by-step answer :
We have to evaluate the limit $\lim \dfrac{{1 - \cos x}}{x}$ as $x \to 0$ using L’Hospital’s rule. So, if we put the limit x tending to zero into the expression $\dfrac{{1 - \cos x}}{x}$ , we get an indeterminate form limit. Hence, L’Hospital’s rule can be applied here to find the value of the concerned limit.
So, Applying L’Hospital’s rule, we have to differentiate both numerator and denominator with respect to the variable x separately and then apply the limit.
Hence, $\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{1 - \cos x}}{x}} \right)$
Now, the derivative of $1 - \cos x$ can be evaluated as $\sin x$ because we know that differentiation of a constant is zero and differentiation of $\cos x$ is $\left( { - \sin x} \right)$ and the derivative of x is $1$ . Hence, we get,
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\sin x}}{1}} \right)$
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \left( {\sin x} \right)$
Putting x as zero, we get,
$\Rightarrow \sin \left( 0 \right)$
$\Rightarrow 0$
So, the value of the limit $\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{1 - \cos x}}{x}} \right)$ is $0$ .
So, the correct answer is “0”.

Note : The given question can also be solved in a different method. We must remember some basic trigonometric formulae to solve the limit through another method. First, we would have to convert the expression $\left( {1 - \cos x} \right)$ in terms of $\sin \dfrac{x}{2}$ and then substitute the value of some basic limits to find the final answer.