\(\mathop {\lim }\limits_{x \to 0} \frac{{1 - {{\cos }^3}4x}}{{{x^2}}}\) is equal to

Concept:

L’ Hospital’s rule:

If \(\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{0}{0}\) then we have to differentiate both the numerator and denominator with respect to x unless and until \(\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = l \ne \frac{0}{0}\) where l is a finite value.

Calculation:

Given: \(\mathop {\lim }\limits_{x \to 0} \frac{{1 - {{\cos }^3}4x}}{{{x^2}}}\)

\(\Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{1 - {{\cos }^3}4x}}{{{x^2}}} = \frac{0}{0}\)

By applying L hospital’s rule we get

\(\Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{1 - {{\cos }^3}4x}}{{{x^2}}} = \;\mathop {\lim }\limits_{x\; \to 0} \frac{{12{{\cos }^2}4x\sin 4x}}{{2x}}\)

\(\Rightarrow \mathop {\lim }\limits_{x\; \to 0} \frac{{12{{\cos }^2}4x\sin 4x}}{{2x}} = \;\mathop {\lim }\limits_{x \to 0} \frac{{24\;\left( {{{\cos }^2}4x\sin 4x} \right)}}{{4x}}\)

\(\Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{24\;\left( {{{\cos }^2}4x\sin 4x} \right)}}{{4x}} = 24 \times \;\mathop {\lim }\limits_{x \to 0} {\cos ^2}4x \times \;\mathop {\lim }\limits_{x \to 0} \frac{{\sin 4x}}{{4x}}\)