JEE MAIN - Mathematics (2023 - 31st January Evening Shift - No. 13)
$$
\lim\limits_{x \rightarrow \infty} \frac{(\sqrt{3 x+1}+\sqrt{3 x-1})^6+(\sqrt{3 x+1}-\sqrt{3 x-1})^6}{\left(x+\sqrt{x^2-1}\right)^6+\left(x-\sqrt{x^2-1}\right)^6} x^3
$$
is equal to 9
is equal to $\frac{27}{2}$
does not exist
is equal to 27
Explanation
$\lim \limits_{x \rightarrow \infty} \frac{(\sqrt{3 x+1}+\sqrt{3 x-1})^{6}+(\sqrt{3 x+1}-\sqrt{3 x-1})^{6}}{\left(x+\sqrt{x^{2}-1}\right)^{6}+\left(x-\sqrt{x^{2}-1}\right)^{6}} x^{3}$
$$ \begin{aligned} & = \lim \limits_{x \rightarrow \infty} x^{3} \times\left\{\frac{x^{3}\left\{\left(\sqrt{3+\frac{1}{x}}+\sqrt{3-\frac{1}{x}}\right)^{6}+\left(\sqrt{3+\frac{1}{x}}-\sqrt{3-\frac{1}{x}}\right)^{6}\right\}}{x^{6}\left\{\left(1+\sqrt{1-\frac{1}{x^{2}}}\right)^{6}+\left(1-\sqrt{1-\frac{1}{x^{2}}}\right)^{6}\right\}}\right\} \\\\ & =\frac{(2 \sqrt{3})^{6}+0}{2^{6}+0}=3^{3}=27 \end{aligned} $$
$$ \begin{aligned} & = \lim \limits_{x \rightarrow \infty} x^{3} \times\left\{\frac{x^{3}\left\{\left(\sqrt{3+\frac{1}{x}}+\sqrt{3-\frac{1}{x}}\right)^{6}+\left(\sqrt{3+\frac{1}{x}}-\sqrt{3-\frac{1}{x}}\right)^{6}\right\}}{x^{6}\left\{\left(1+\sqrt{1-\frac{1}{x^{2}}}\right)^{6}+\left(1-\sqrt{1-\frac{1}{x^{2}}}\right)^{6}\right\}}\right\} \\\\ & =\frac{(2 \sqrt{3})^{6}+0}{2^{6}+0}=3^{3}=27 \end{aligned} $$
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