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JEE Mathematics 2025 Question with Solution

If lim⁡x→∞(e1−e(1e−x1+x))x=α,\lim_{x \to \infty} \left( \frac{e}{1 - e} \left( \frac{1}{e} - \frac{x}{1 + x} \right) \right)^x = \alpha, then the value of log⁡eα1+log⁡eα\frac{\log_e \alpha}{1 + \log_e \alpha} equals:

  • A

    e−2e^{-2}

  • B

    e−1e^{-1}

  • C

    ee

  • D

    e2e^2

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given:

α=lim⁡x→∞(e1−e(1e−x1+x))x\alpha = \lim_{x \to \infty} \left( \frac{e}{1-e}\left(\frac{1}{e}-\frac{x}{1+x}\right) \right)^x

Find:

log⁡eα1+log⁡eα\frac{\log_e \alpha}{1+\log_e \alpha}

First simplify the factor inside the bracket:

x1+x=1−11+x\frac{x}{1+x} = 1 - \frac{1}{1+x}

So,

1e−x1+x=1e−1+11+x=1−ee+11+x\frac{1}{e} - \frac{x}{1+x} = \frac{1}{e} - 1 + \frac{1}{1+x} = \frac{1-e}{e} + \frac{1}{1+x}

Multiply by e1−e\frac{e}{1-e}:

e1−e(1−ee+11+x)=1+e(1−e)(1+x)\frac{e}{1-e}\left(\frac{1-e}{e} + \frac{1}{1+x}\right)=1+\frac{e}{(1-e)(1+x)}

Hence,

α=lim⁡x→∞(1+e(1−e)(1+x))x\alpha = \lim_{x \to \infty} \left(1+\frac{e}{(1-e)(1+x)}\right)^x

Now use the standard limit form (1+cx)x→ec\left(1+\frac{c}{x}\right)^x \to e^c. Since

e(1−e)(1+x)∼e(1−e)x\frac{e}{(1-e)(1+x)} \sim \frac{e}{(1-e)x}

as x→∞x \to \infty, we get

α=ee1−e\alpha = e^{\frac{e}{1-e}}

Therefore,

log⁡eα=e1−e\log_e \alpha = \frac{e}{1-e}

So,

log⁡eα1+log⁡eα=e1−e1+e1−e=e1−e11−e=e\frac{\log_e \alpha}{1+\log_e \alpha} = \frac{\frac{e}{1-e}}{1+\frac{e}{1-e}} = \frac{\frac{e}{1-e}}{\frac{1}{1-e}} = e

Therefore, the required value is ee. The mathematically consistent result is option C. The solution contains conflicting and incorrect conclusions, but the limit evaluation gives ee.

Using logarithm of the limit

Write

α=lim⁡x→∞(1+ux)x\alpha = \lim_{x\to\infty} \left(1+u_x\right)^x

where

ux=e(1−e)(1+x)u_x = \frac{e}{(1-e)(1+x)}

Then

log⁡α=lim⁡x→∞xlog⁡(1+ux)\log \alpha = \lim_{x\to\infty} x\log(1+u_x)

Since ux→0u_x \to 0, use log⁡(1+ux)∼ux\log(1+u_x) \sim u_x:

log⁡α=lim⁡x→∞x⋅e(1−e)(1+x)=e1−e\log \alpha = \lim_{x\to\infty} x \cdot \frac{e}{(1-e)(1+x)} = \frac{e}{1-e}

Hence,

α=ee1−e\alpha = e^{\frac{e}{1-e}}

Now substitute into the required expression:

log⁡eα1+log⁡eα=e1−e1+e1−e=e\frac{\log_e \alpha}{1+\log_e \alpha} = \frac{\frac{e}{1-e}}{1+\frac{e}{1-e}} = e

So the correct option is C.

Common mistakes

  • Treating x1+x\frac{x}{1+x} as exactly 11 too early. That loses the small term of order 1x\frac{1}{x}, which is crucial in a limit of the form (1+cx)x\left(1+\frac{c}{x}\right)^x. Simplify up to the first-order term before applying the standard exponential limit.

  • Using the provided option or claimed answer without checking the algebra. The solution itself is inconsistent. Always derive α\alpha from the expression first, then evaluate the asked quantity.

  • Miscomputing

    e1−e1+e1−e\frac{\frac{e}{1-e}}{1+\frac{e}{1-e}}

    by not combining the denominator correctly. Convert the denominator to a single fraction first; it becomes 11−e\frac{1}{1-e}, which leads to the final result ee.

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