What is the value of ${lim}_{x \to 0} \left( {\dfrac{d}{{dx}}} \right)\left[ {\int {\dfrac{{\left( {1 - \cos x} \right)}}{{{x^2}}}} } \right]dx$?
A. 1
B. 0
C. $\dfrac{1}{2}$
D. None of these

Hint: Cancel out derivative and integral, because in calculus integral is opposite of derivative. Then apply the L’Hospital rule. Then substitute the given values of the variable in the limit and get the required answer.

Formula Used:
L’Hospital rule:
If $\dfrac{{f\left( x \right)}}{{g\left( x \right)}}$ is in the form $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$ when $x \to c$, then ${lim}_{x \to c} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = {lim}_{x \to c} \dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}$.
${lim}_{x \to 0} \left[ {\dfrac{{\sin x}}{x}} \right] = 1$

Complete step by step solution:
The given limit is ${lim}_{x \to 0} \left( {\dfrac{d}{{dx}}} \right)\left[ {\int {\dfrac{{\left( {1 - \cos x} \right)}}{{{x^2}}}} } \right]dx$.
Let’s solve the given limit.
Let $L$ be the value of the limit.
$L = {lim}_{x \to 0} \left( {\dfrac{d}{{dx}}} \right)\left[ {\int {\dfrac{{\left( {1 - \cos x} \right)}}{{{x^2}}}} } \right]dx$
Since the opposite of the derivative is integral.
So, they cancel each other out.
$L = {lim}_{x \to 0} \left[ {\dfrac{{\left( {1 - \cos x} \right)}}{{{x^2}}}} \right]$
The right-hand side of the limit is in the form of $\dfrac{0}{0}$.
So, apply the L’Hospital rule.
Differentiate the right-hand side of the limit with respect to $x$.
$L = {lim}_{x \to 0} \left[ {\dfrac{{\left( {0 - \left( { - \sin x} \right)} \right)}}{{2x}}} \right]$ [Since $\dfrac{d}{{dx}}\cos x = - \sin x$ and $\dfrac{d}{{dx}}{x^n} = n{x^{\left( {n - 1} \right)}}$]
$ \Rightarrow L = {lim}_{x \to 0} \left[ {\dfrac{{\sin x}}{{2x}}} \right]$
$ \Rightarrow L = \dfrac{1}{2}{lim}_{x \to 0} \left[ {\dfrac{{sin x}}{x}} \right]$
$ \Rightarrow L = \dfrac{1}{2} \times 1$ [Since ${lim}_{x \to 0} \left[ {\dfrac{{\sin x}}{x}} \right] = 1$]
$ \Rightarrow L = \dfrac{1}{2}$

Option ‘C’ is correct

Note: When the numerator and denominator of a limit are in the form of $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$, we use L’Hospital rule. In this process, we take derivatives of numerator and denominator with respect to the given variable. So that the function is no longer in the indeterminate form.