Q105 / 160 · TS EAMCET 2022 (19 Jul Shift 1) · Mathematics

$$ \lim _{x \rightarrow 0} \frac{\tan ^2\left(\pi \sec ^4 x\right)}{\pi^2 x^4}= $$

  • A
    0
  • Correct answer
    B
    4
  • C
    1
  • D
    16

Solution

Given $\lim _{x \rightarrow 0} \frac{\tan ^2\left(\pi \sec ^4 x\right)}{\pi^2 x^2}$ It forms $\frac{0}{0}$ indeterminate form, then apply L'Hospital rules. $$ \begin{aligned} \Rightarrow & \lim _{x \rightarrow 0} \frac{\tan ^2\left(\pi \sec ^4 x\right)}{\pi^2 x^4} \\ & \lim _{x \rightarrow 0} \frac{2 \tan \left(\pi \sec ^4 x\right) \cdot \sec ^2\left(\pi \sec ^4 x\right) \cdot 4 \pi \sec ^4 x \tan x}{4 \pi^2 x^3} \\ \Rightarrow & \lim _{x \rightarrow 0}\left(\frac{\left.2 \tan \left(\pi \sec ^4 x\right) \cdot \sec ^2\left(\pi \sec ^4 x\right) \cdot \sec ^4 x\right)}{\pi x^2}\right) \cdot \lim _{x \rightarrow 0}\left(\frac{\tan x}{x}\right) \\ \Rightarrow & \lim _{x \rightarrow 0}\left(\frac{2 \tan \left(\pi \sec ^4 x\right)}{x^2}\right) \cdot \lim _{x \rightarrow 0}\left(\frac{\sec ^2\left(\pi \sec ^4 x\right) \cdot \sec ^4 x}{\pi}\right) \cdot 1 \end{aligned} $$ Again Apply L'Hospital rule, $$ \begin{aligned} & \Rightarrow \lim _{x \rightarrow 0}\left(\frac{2 \sec ^2\left(\pi \sec ^4 x\right) \cdot 4 \pi \sec ^4 x \tan x}{2 x}\right) \frac{1}{\pi}\left(\sec ^2\left(\pi \sec ^4 0\right) \cdot \sec ^4(0)\right) \\ & \Rightarrow \lim _{x \rightarrow 0}\left(\frac{\tan x}{x}\right) \cdot \lim _{x \rightarrow 0} \frac{\left(\sec ^2\left(\pi \sec ^4 x\right) \cdot 4 x \sec ^4 x\right)}{\pi} \cdot\left(\sec ^2(\pi) \cdot 1\right) \\ & \quad \Rightarrow 1 \cdot \sec ^2\left(\pi \sec ^4(0)\right) \cdot 4 \cdot \sec ^4(0) \cdot(-1)^2 \\ & \quad \Rightarrow 1 \cdot(-1)^2 \cdot 4 \cdot(1)=4 \end{aligned} $$

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

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