If $\lim_{x\rightarrow \infty}\left(\left(\frac{e}{1-e}\right)\left(\frac{1}{e}-\frac{x}{1+x}\right)\right)^{x} = \alpha$, then the value of $\frac{\log_{e}\alpha}{1+\log_{e}\alpha}$ equals:
- (1) e
- (2) $e^{-2}$
- (3) $e^{2}$
- (4) $e^{-1}$
Solution:
The limit is of the form $1^{\infty}$ as $x \to \infty$:
Inside term: $\frac{e}{1-e} \left( \frac{1}{e} – 1 \right) = \frac{e}{1-e} \left( \frac{1-e}{e} \right) = 1$.
$\alpha = e^L$
Let $L = \lim_{x\rightarrow \infty} x \left[ \left(\frac{e}{1-e}\right)\left(\frac{1}{e}-\frac{x}{1+x}\right) – 1 \right]$.
Simplify the term in the bracket:
$$ \left(\frac{e}{1-e}\right)\left(\frac{1+x-ex}{e(1+x)}\right) – 1 = \frac{1+x-ex}{(1-e)(1+x)} – 1 $$
$$ = \frac{1+x-ex – (1-e)(1+x)}{(1-e)(1+x)} $$
$$ = \frac{1+x-ex – (1+x-e-ex)}{(1-e)(1+x)} $$
$$ = \frac{e}{(1-e)(1+x)} $$
Now, substitute back into the limit:
$$ L = \lim_{x\rightarrow \infty} x \cdot \frac{e}{(1-e)(1+x)} = \frac{e}{1-e} \lim_{x\rightarrow \infty} \frac{x}{1+x} $$
$$ L = \frac{e}{1-e} (1) = \frac{e}{1-e} $$
Thus, $\alpha = e^L = e^{\frac{e}{1-e}}$.
Taking log: $\log_e \alpha = \frac{e}{1-e}$.
We need to find $\frac{\log_e \alpha}{1 + \log_e \alpha}$:
$$ \frac{\frac{e}{1-e}}{1 + \frac{e}{1-e}} = \frac{\frac{e}{1-e}}{\frac{1-e+e}{1-e}} = \frac{\frac{e}{1-e}}{\frac{1}{1-e}} = e $$
Ans. (1)