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Limits Continuity and Differentiability question

2025 · 23 Jan · Shift 2 · Q37

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Limits Continuity and Differentiability
2025 · 23 Jan · Shift 2 · Q37

Limits Continuity and Differentiability question

2025 · 23 Jan · Shift 2 · Q37

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1

\lim \limits_{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 :

A
$\frac{2 e}{3}$
B
$\frac{2}{3 \sqrt{\mathrm{e}}}$
C
$\frac{2 \mathrm{e}}{\sqrt{3}}$
D
$\frac{2}{\sqrt{3 \mathrm{e}}}$

View written solutionFree

Correct answer: B

Write the limit in a simpler form

We need to evaluate

L=\lim_{x\to\infty}\frac{(2x^2-3x+5)(3x-1)^{x/2}}{(3x^2+5x+4)\sqrt{(3x+2)^x}}.

Since

\sqrt{(3x+2)^x}=(3x+2)^{x/2},

we get

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

So the limit is the product of two simpler limits.

Evaluate the rational factor

\lim_{x\to\infty}\frac{2x^2-3x+5}{3x^2+5x+4}=\frac{2}{3}.

Evaluate the exponential factor

Consider

M=\lim_{x\to\infty}\left(\frac{3x-1}{3x+2}\right)^{x/2}.

First simplify the base:

\frac{3x-1}{3x+2}=1-\frac{3}{3x+2}.

Hence

M=\lim_{x\to\infty}\left(1-\frac{3}{3x+2}\right)^{x/2}.

Now use the standard exponential-limit form. Let

\ln M=\lim_{x\to\infty}\frac{x}{2}\ln\left(1-\frac{3}{3x+2}\right).

Using $\ln(1+u)\sim u$ as $u\to 0$ ,

\ln M=\lim_{x\to\infty}\frac{x}{2}\left(-\frac{3}{3x+2}\right) =-\frac{1}{2}\lim_{x\to\infty}\frac{3x}{3x+2}=-\frac{1}{2}.

Therefore

M=e^{-1/2}=\frac{1}{\sqrt e}.

Combine both limits

Thus,

L=\frac{2}{3}\cdot\frac{1}{\sqrt e}=\frac{2}{3\sqrt e}.

Match with the options

\boxed{\frac{2}{3\sqrt e}}

which is Option B.

Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So they agree.

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