Q102 / 160 · TS EAMCET 2018 (05 May Shift 2) · Mathematics

$\lim _{x \rightarrow 0} \frac{\cos 4 x-4 \cos 2 x+3}{x^4}=$

  • A
    4
  • Correct answer
    B
    8
  • C
    $\frac{1}{4}$
  • D
    $\frac{1}{8}$

Solution

We have, $$ \begin{aligned} & \lim _{x \rightarrow 0} \frac{\cos 4 x-4 \cos 2 x+3}{x^4} \quad\left[\frac{0}{0} \text { form }\right] \\ & =\lim _{x \rightarrow 0} \frac{-4 \sin 4 x+8 \sin 2 x}{4 x^3} \quad\left[\frac{0}{0} \text { form }\right] \\ & =\lim _{x \rightarrow 0} \frac{-16 \cos 4 x+16 \cos 2 x}{12 x^2} \quad\left[\frac{0}{0} \text { form }\right] \\ & \end{aligned} $$ $$ \begin{array}{ll} =\lim _{x \rightarrow 0} \frac{+64 \sin 4 x-32 \sin 2 x}{24 x} & \quad\left[\frac{0}{0} \text { form }\right] \\ =\lim _{x \rightarrow 0} \frac{256 \cos 4 x-64 \cos 2 x}{24} & \\ =\frac{256-64}{24}=\frac{192}{24}=8 & \end{array} $$

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

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