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BITSAT 2011
Exam
BITSAT
Year
2011
Subject
Mathematics
Chapter
Limits

BITSAT 2011 · Mathematics · Limits

Evaluate ( _ x arrow ∞ 1+ 3 x -1 (1+ 2 x) );

Evaluate (\lim _{x \rightarrow \infty} \frac{\sqrt{1+\sin 3 x}-1}{\ln (1+\tan 2 x)});

Options

  1. A1/21 / 2
  2. B3/23 / 2
  3. C3/43 / 4
  4. D1/41 / 4

Correct answer

C. 3/43 / 4

Step-by-step solution

(lim⁡x→01+sin⁡3x−1ln⁡(1+tan⁡2x)=lim⁡x→0(1+sin⁡3x)−1{1+sin⁡3x+1)ln⁡(1+tan⁡2x)=lim⁡x→01(1+sin⁡3x+1)⋅sin⁡3xln⁡(1+tan⁡2x)1tan⁡2x×1tan⁡2x\begin{aligned} & \lim _{x \rightarrow 0} \frac{\sqrt{1+\sin 3 x}-1}{\ln (1+\tan 2 x)} \\ = & \lim _{x \rightarrow 0} \frac{(1+\sin 3 x)-1}{\{\sqrt{1+\sin 3 x+1}) \ln (1+\tan 2 x)} \\ = & \lim _{x \rightarrow 0} \frac{1}{(\sqrt{1+\sin 3 x}+1)} \cdot \frac{\sin 3 x}{\ln (1+\tan 2 x)^{\frac{1}{\tan 2 x}}\times \frac{1}{\tan 2 x}} \end{aligned}) (=lim⁡x→01(1+sin⁡3x+1)⋅(sin⁡3x3x)(2xtan⁡2x)×3tan⁡2x⋅1ln⁡(1+tan⁡2x)1tan⁡2x=(11+1)(1)(1)(32)1ln⁡e=34.\begin{aligned} & =\lim _{x \rightarrow 0} \frac{1}{(\sqrt{1+\sin 3 x}+1)} \cdot\left(\frac{\sin 3 x}{3 x}\right)\left(\frac{2 x}{\tan 2 x}\right) \\ & \times \frac{3}{\tan 2 x} \cdot \frac{1}{\ln (1+\tan 2 x)^{\frac{1}{\tan 2 x}}} \\ & =\left(\frac{1}{1+1}\right)(1)(1)\left(\frac{3}{2}\right) \frac{1}{\ln e}=\frac{3}{4} .\end{aligned})
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