🛠️ JEE➗ Maths

\(\lim _{x \rightarrow \infty} \frac{\left(2 x^2-3 x+5\right)(3 x-1)^{\frac{x}{2}}}{\left(3 x^2+5 x+4\right) \sqrt{(3 x+…

Q1 FREE PREVIEW

\(\lim _{x \rightarrow \infty} \frac{\left(2 x^2-3 x+5\right)(3 x-1)^{\frac{x}{2}}}{\left(3 x^2+5 x+4\right) \sqrt{(3 x+2)^x}}\) is equal to

[JEE Main 2025, 23 Jan (Shift 2)]

a

\(\frac{2}{\sqrt{3 \mathrm{e}}}\)

b

\(\frac{2 \mathrm{e}}{\sqrt{3}}\)

c

\(\frac{2 \mathrm{e}}{3}\)

d

\(\frac{2}{3 \sqrt{e}}\)

✓ Correct answer: d)

\(\frac{2}{3 \sqrt{e}}\)

Explanation

\(\lim _{x \rightarrow \infty} \frac{\left(2 x^2-3 x+5\right)(3 x-1)^{\frac{x}{2}}}{\left(3 x^2+5 x+4\right) \sqrt{(3 x+2)^x}}\)

\(=\lim _{x\to \infty }\frac{\left(2-\frac{3}{x}+\frac{5}{{x}^{2}}\right){\left(1-\frac{1}{3x}\right)}^{x/2}}{\left(3+\frac{5}{x}+\frac{4}{{x}^{2}}\right){\left(1+\frac{2}{3x}\right)}^{x/2}}\)

\(=\lim _{x\to \infty }\frac{\left(2-\frac{3}{x}+\frac{5}{{x}^{2}}\right)}{\left(3+\frac{5}{x}+\frac{4}{{x}^{2}}\right)}\lim _{x\to \infty }\frac{{\left(1-\frac{1}{3x}\right)}^{x/2}}{{\left(1+\frac{2}{3x}\right)}^{x/2}}\)

\(=\frac{2}{3}\lim _{x\to \infty }{\left(\frac{\left(1-\frac{1}{3x}\right)}{\left(1+\frac{2}{3x}\right)}\right)}^{x/2}\)

\(=\frac{2}{3}\cdot \frac{{e}^{\frac{x}{2}\left(1-\frac{1}{3x}-1\right)}}{{e}^{\frac{x}{2}\left(1+\frac{2}{3x}-1\right)}}\)

\(=\frac{2}{3}\cdot \frac{{\mathrm{e}}^{-\frac{1}{6}}}{{\mathrm{e}}^{1/3}}=\frac{2}{3}{\mathrm{e}}^{-\frac{1}{2}}\)

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