Q225 / 270 · NDA 2019 (Phase 1) · Mathematics

$\lim _{x \rightarrow \infty} \frac{1-\cos ^3 4 x}{x^2}$ is equal to

  • A
    0
  • B
    12
  • Correct answer
    C
    24
  • D
    36

Solution

$\lim _{x \rightarrow 0} \frac{1-\cos ^3 4 x}{x^2}$ The given limit is in $\frac{0}{0}$ form. So, apply L'hospital rule. $\begin{aligned} & \lim _{x \rightarrow 0} \frac{-3 \cos ^2 4 x(-\sin 4 x) 4}{2 x} \\ & =\lim _{x \rightarrow 0} \frac{12 \cos ^2 4 x \sin 4 x}{(2 x) 2} \times 2 \\ & =24 \cdot \lim _{x \rightarrow 0} \cos ^2 4 x \cdot \frac{\sin 4 x}{4 x} \\ & =24(1)=24\end{aligned}$

Hence, the correct answer is option: 'C'.

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