Medium MCQ +4 / -1 PYQ · JEE Mains 2023

$$ \lim\limits_{x \rightarrow \infty} \frac{(\sqrt{3 x+1}+\sqrt{3 x-1})^6+(\sqrt{3 x+1}-\sqrt{3 x-1})^6}{\left(x+\sqrt{x^2-1}\right)^6+\left(x-\sqrt{x^2-1}\right)^6} x^3 $$

  1. A is equal to 9
  2. B is equal to $\frac{27}{2}$
  3. C does not exist
  4. D is equal to 27 Correct answer

Solution

$\lim \limits_{x \rightarrow \infty} \frac{(\sqrt{3 x+1}+\sqrt{3 x-1})^{6}+(\sqrt{3 x+1}-\sqrt{3 x-1})^{6}}{\left(x+\sqrt{x^{2}-1}\right)^{6}+\left(x-\sqrt{x^{2}-1}\right)^{6}} x^{3}$ <br/><br/>$$ \begin{aligned} & = \lim \limits_{x \rightarrow \infty} x^{3} \times\left\{\frac{x^{3}\left\{\left(\sqrt{3+\frac{1}{x}}+\sqrt{3-\frac{1}{x}}\right)^{6}+\left(\sqrt{3+\frac{1}{x}}-\sqrt{3-\frac{1}{x}}\right)^{6}\right\}}{x^{6}\left\{\left(1+\sqrt{1-\frac{1}{x^{2}}}\right)^{6}+\left(1-\sqrt{1-\frac{1}{x^{2}}}\right)^{6}\right\}}\right\} \\\\ & =\frac{(2 \sqrt{3})^{6}+0}{2^{6}+0}=3^{3}=27 \end{aligned} $$

About this question

Subject: Mathematics · Chapter: Limits, Continuity and Differentiability · Topic: Limits and Standard Results

This question is part of PrepWiser's free JEE Main question bank. 162 more solved questions on Limits, Continuity and Differentiability are available — start with the harder ones if your accuracy is >70%.

Drill 25 more like these. Every day. Free.

PrepWiser turns these solved questions into a daily practice loop. Chapter-wise drills, full mocks, AI doubt chat. No auto-renew.

Start free →