Q101 / 160 · TS EAMCET 2024 (09 May Shift 2) · Mathematics

$\lim _{\theta \rightarrow \frac{\pi}{2}^{-}} \frac{8 \tan ^4 \theta+4 \tan ^2 \theta+5}{(3-2 \tan \theta)^4}=$

  • A
    $-\frac{1}{2}$
  • Correct answer
    B
    $\frac{1}{2}$
  • C
    -4
  • D
    1

Solution

$\begin{aligned} & \lim _{\theta \rightarrow \frac{\pi^{-}}{2}} \frac{8 \tan ^4 \theta+4 \tan ^2 \theta+5}{(3-2 \tan \theta)^4} \quad \mathrm{[Using L'Hopital]} \\ & =\lim _{\theta \rightarrow \frac{\pi^{-}}{2}} \frac{\left(4 \tan ^2 \theta+1\right) \tan \theta}{(2 \tan \theta-3)^3} \quad \mathrm{[Using L'Hopital]} \end{aligned}$ $\begin{aligned} & =\lim _{\theta \rightarrow \frac{\pi^{-}}{2}} \frac{12 \tan ^2 \theta+1}{6(2 \tan \theta-3)^2} \qquad \mathrm{[Using L'Hopital]} \\ & =\lim _{\theta \rightarrow \frac{\pi^{-}}{2}} \frac{\tan \theta}{2 \tan \theta-3}=\frac{1}{2} \qquad \mathrm{[Using L'Hopital]} \end{aligned}$

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

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