Animated Solution for Mathematics - Limits, Continuity and Differentiability: If ${\lim }_{x\to \infty }((\frac{e}{1-e})(\frac{1}{e}-\frac{x}{1+x}){)}^x=\alpha$, then the value of $\frac{{\log }_e\alpha }{1+{\log }_e\alpha }$ equals

Imagine you are staring at this limit. It looks like a monster, doesn't it? A complex fraction raised to the power of $x$, with $x$ running off to infinity.

In the world of JEE Advanced, when you see a variable in the base and a variable in the exponent, your first instinct should be to check the form. Is it $0^0$? Is it ${\infty }^0$? Or is it the classic, the legendary $1^{\infty }$?

Let us dissect the base first. We have:

As $x\to \infty$, the term $\frac{x}{1+x}$ behaves like $\frac{x}{x}$, which is $1$. So, our base becomes:

If you simplify that, $\left(\frac{1}{e}-1\right)$ is $\left(\frac{1-e}{e}\right)$. When you multiply $\left(\frac{e}{1-e}\right)$ by $\left(\frac{1-e}{e}\right)$, everything cancels out to $1$. We have confirmed it: the base is $1$, and the exponent is $\infty$. We are in the territory of the $1^{\infty }$ indeterminate form.

Now, do not panic. We do not need to take logarithms and perform complex differentiation. We have a standard, elegant formula for this:

This formula is your best friend. It takes that terrifying exponent and brings it down to the ground, turning a power problem into a simple multiplication problem. Let us define $L={\lim }_{x\to \infty }x[f(x)-1]$. This $L$ is the key to the kingdom.

This is where the real work happens. We need to compute:

Let us focus on the bracket. We need a common denominator. The expression inside is:

If we take the common denominator $(1-e)(1+x)$, the numerator becomes $(1+x)-ex-(1-e)(1+x)$. Now, watch the magic. Expand that numerator:

The $1$ cancels with $-1$, the $x$ cancels with $-x$, and the $-ex$ cancels with $+ex$. We are left with only $e$ in the numerator! Our massive, scary bracket has reduced to:

Now, bring back the $x$ we left outside. We have:

We can pull the constant $\frac{e}{1-e}$ outside the limit. We are left with ${\lim }_{x\to \infty }\frac{x}{1+x}$, which is clearly $1$. Thus:

Since $\alpha =e^L$, taking the natural log gives us ${\log }_e\alpha =L=\frac{e}{1-e}$. The final step is just arithmetic. We need to evaluate:

Substituting our value, we get:

The denominator becomes $\frac{1-e+e}{1-e}=\frac{1}{1-e}$. When you divide the numerator by this denominator, the $(1-e)$ terms vanish, and we are left with exactly $e$. A complex, intimidating limit, reduced to a single, beautiful constant. That is the power of systematic thinking.