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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Limits

NTA Abhyas 2020 · Mathematics · Limits

The value of _ x → 0 tan^ 2 3x 5 - 4 + x & is equal to

The value of lim⁡x→0tan23x5−4+sec⁡x\lim_{x \to 0}\dfrac{tan^{2}3x}{\sqrt{5} - \sqrt{4 + \sec x}}& is equal to

Options

  1. A252\sqrt{5}
  2. B−95- 9\sqrt{5}
  3. C959\sqrt{5}
  4. D−365- 36\sqrt{5}

Correct answer

D. −365- 36\sqrt{5}

Step-by-step solution

After rationalization,
& lim⁡x→0tan23x(5+4+sec⁡x)5−(4+secx)\lim_{x \to 0}\dfrac{tan^{2}3x\left(\sqrt{5} + \sqrt{4 + \sec x}\right)}{5 - \left(4 + sec x\right)}
⇒lim⁡x→0(tan23x9x2)⇒ \lim_{x \to 0} \left(\dfrac{tan^{2}3x}{9x^{2}}\right)⋅ 9x^{2}\left(\cos x - 1\right)\left(\sqrt{5} + \sqrt{4 + 1}\right)
⇒9(−2)(25)⇒ 9 \left(-2\right)\left(2\sqrt{5}\right) = -36\sqrt{5}
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