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

(D) is equal to $\frac{27}{2}$

JEE Main · 2023 · 31 Jan 2023 · Shift 2 · Limits

lim_{x \to \infty} \frac{( 3 x + 1 + 3 x - 1 ) ^{6} + ( 3 x + 1 - 3 x - 1 ) ^{6}}{( x + x ^{2} - 1 ) ^{6} + ( x - x ^{2} - 1 ) ^{6}} x^{3}

Answer: (B)

x \to \infty

3 x + 1 + 3 x - 1 \to 2 3 x

while

3 x + 1 - 3 x - 1 = \frac{2}{3 x + 1 + 3 x - 1} \to 0

, so the numerator bracket behaves like

(2 3 x)^{6} = 1728 x^{3}

, and multiplying by

x^{3}

gives

1728 x^{6} .

Likewise

x + x^{2} - 1 \to 2 x

and

x - x^{2} - 1 \to 0

, so the denominator behaves like

(2 x)^{6} = 64 x^{6} .

The quotient therefore tends to

\frac{1728}{64} = 27.

JEE Main · 2023 · 31 Jan 2023 · Shift 2 · Single correct option

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