Question

If $\lim_{x \to \infty} ((\frac{e}{1-e})-(\frac{1}{1+x}))^{x} = a$, then the value of $\frac{\log_{e} \alpha}{1+\log_{e} \alpha}$ equals:

• A. $e^{2}$

Answer 1

e^{2}

Answer 2

The given limit is $L = \lim_{x \to \infty} \left(\left(\frac{e}{1-e}\right)-\left(\frac{1}{1+x}\right)\right)^{x}$. Let the base of the expression be $B(x) = \frac{e}{1-e} - \frac{1}{1+x}$. As $x \to \infty$, the term $\frac{1}{1+x} \to 0$. So, the base $B(x)$ approaches a constant value $C = \frac{e}{1-e}$.

We know that $e \approx 2.718$. Therefore, $1-e \approx 1 - 2.718 = -1.718$. So, $C = \frac{e}{1-e} \approx \frac{2.718}{-1.718} \approx -1.58$.

The limit is of the form $\lim_{x \to \infty} (C)^x$, where $C \approx -1.58$. Since $C < -1$, the limit $\lim_{x \to \infty} C^x$ does not exist in real numbers. For example, if $x$ takes integer values, $C^x$ oscillates between positive and negative values, and its magnitude tends to infinity. For example, if $C=-2$, then $(-2)^1=-2$, $(-2)^2=4$, $(-2)^3=-8$, etc.

If the limit $a$ does not exist as a real number, then $\log_e a$ is undefined. This implies that the question as stated is problematic.

However, in competitive exams, if such a question appears, it often implies a typo, and the intended form is usually a standard limit of the form $1^\infty$. For a limit of the form $\lim_{x \to \infty} (f(x))^{g(x)}$ to be of the $1^\infty$ indeterminate form, we must have $\lim_{x \to \infty} f(x) = 1$ and $\lim_{x \to \infty} g(x) = \infty$. In our case, $g(x) = x \to \infty$. For $f(x)$ to approach 1, we would need $\frac{e}{1-e} = 1$, which implies $e = 1-e$, or $2e=1$, so $e=1/2$. This is false.

Given the common types of limits and the options provided (which are real numbers), let's assume there was a typo and the intended form is usually a standard limit of the form $1^\infty$.

The problem as stated leads to a limit that does not exist in real numbers. If it is a complex limit, then it is beyond the scope of typical JEE/NEET syllabus.

Final conclusion based on the strict interpretation of the question: The limit does not exist, so $\log_e \alpha$ is undefined.

Given the constraints, the question as written is invalid for real numbers.