JEE Main 2025 Jan 22 Shift 2, Mathematics Q7: Limits, Continuity And Differentiability

As $x\to \infty$, $\frac{1}{e}-\frac{x}{1+x}\to \frac{1}{e}-1=\frac{1-e}{e}$, and the prefactor $\frac{e}{1-e}$ multiplied by this tends to $1$. So we have a $1^{\infty }$ form, handled by $\alpha =e^L$ where:

$L={\lim }_{x\to \infty }x\left[\frac{e}{1-e}\left(\frac{1}{e}-\frac{x}{1+x}\right)-1\right]$.

Simplify the bracket. Note $\frac{e}{1-e}\cdot \frac{1}{e}=\frac{1}{1-e}$, and $-1=\frac{e-1}{1-e}\cdot (-1)=\frac{1-e}{1-e}\cdot (-1)$, so:

$\frac{e}{1-e}\left(\frac{1}{e}-\frac{x}{1+x}\right)-1=\frac{1}{1-e}-\frac{e}{1-e}\cdot \frac{x}{1+x}-1=\frac{e}{1-e}\left(\frac{1}{e}-\frac{x}{1+x}-\frac{1-e}{e}\right)=\frac{e}{1-e}\left(1-\frac{x}{1+x}\right)=\frac{e}{1-e}\cdot \frac{1}{1+x}$.

Therefore $L={\lim }_{x\to \infty }x\cdot \frac{e}{1-e}\cdot \frac{1}{1+x}=\frac{e}{1-e}$.

So $\log \alpha =L=\frac{e}{1-e}$.

Required value: $\frac{\log \alpha }{1+\log \alpha }=\frac{\frac{e}{1-e}}{1+\frac{e}{1-e}}=\frac{\frac{e}{1-e}}{\frac{1-e+e}{1-e}}=\frac{\frac{e}{1-e}}{\frac{1}{1-e}}=e$.