MCQMediumJEE 2025Limits

JEE Mathematics 2025 Question with Solution

Evaluate the following limit: limx(2x23x+5)(3x1)x/2(3x2+5x+4)(3x+2)x.\lim_{x \to \infty} \frac{(2x^2 - 3x + 5) \left( 3x - 1 \right)^{x/2}}{(3x^2 + 5x + 4) \sqrt{(3x + 2)^x}}. The value of the limit is:

  • A

    23e\frac{2}{\sqrt{3e}}

  • B

    2e3\frac{2e}{\sqrt{3}}

  • C

    2e3\frac{2e}{3}

  • D

    23e\frac{2}{3\sqrt{e}}

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given:

L=limx(2x23x+5)(3x1)x/2(3x2+5x+4)(3x+2)xL=\lim_{x \to \infty} \frac{(2x^2 - 3x + 5) (3x - 1)^{x/2}}{(3x^2 + 5x + 4) \sqrt{(3x + 2)^x}}

Find: The value of the limit.

Rewrite the radical term as

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

So,

L=limx2x23x+53x2+5x+4(3x13x+2)x/2L=\lim_{x \to \infty} \frac{2x^2 - 3x + 5}{3x^2 + 5x + 4}\left(\frac{3x-1}{3x+2}\right)^{x/2}

Asymptotic Evaluation

For the polynomial factor,

limx2x23x+53x2+5x+4=23\lim_{x \to \infty} \frac{2x^2 - 3x + 5}{3x^2 + 5x + 4} = \frac{2}{3}

Exponential Limit Trick

For the exponential factor,

3x13x+2=133x+2\frac{3x-1}{3x+2}=1-\frac{3}{3x+2}

Hence,

(3x13x+2)x/2=(133x+2)x/2\left(\frac{3x-1}{3x+2}\right)^{x/2} = \left(1-\frac{3}{3x+2}\right)^{x/2}

Using the standard limit (1+ax)bxeab\left(1+\frac{a}{x}\right)^{bx} \to e^{ab}, we get

(133x+2)x/2e1/2=1e\left(1-\frac{3}{3x+2}\right)^{x/2} \to e^{-1/2}=\frac{1}{\sqrt{e}}

Therefore,

L=231e=23eL=\frac{2}{3}\cdot \frac{1}{\sqrt{e}} = \frac{2}{3\sqrt{e}}

So the correct option is D.

Common mistakes

  • Approximating both exponential terms separately by (3x)x/2(3x)^{x/2} and concluding their ratio is 11 is incomplete. The bases differ by lower-order terms, and when raised to power x/2x/2 they produce a nontrivial exponential factor. First combine them as (3x13x+2)x/2\left(\frac{3x-1}{3x+2}\right)^{x/2} and then apply the standard exponential limit.

  • Cancelling only the leading polynomial terms and stopping at 23\frac{2}{3} misses the contribution from the power term. The polynomial part tends to 23\frac{2}{3}, but the exponential part tends to 1e\frac{1}{\sqrt{e}}, so both factors must be multiplied.

  • Using (1+ux)xe\left(1+u_x\right)^x \to e without checking the limit of xuxx u_x leads to the wrong exponent. Here ux=33x+2u_x = -\frac{3}{3x+2}, so xux1x u_x \to -1, and because the power is x/2x/2, the limit becomes e1/2e^{-1/2}, not e1e^{-1} or 11.

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