Limits, Continuity & Differentiability
$1^\infty$ Form Limits
nta_pyq_2025_apr
Grade 12
Question:
If $\lim_{x \to \infty}\!\left(\!\left(\frac{e}{1-e}\right)\!\left(\frac{1}{e} - \frac{x}{1+x}\right)\!\right)^x = \alpha$, then the value of $\dfrac{\log_e \alpha}{1 + \log_e \alpha}$ equals:
$e^{-1}$
$e^2$
$e^{-2}$
$e$
Step-by-Step Solution
Key Concept: This is a $1^\infty$ form. Take logarithm: $\log\alpha = \lim_{x\to\infty} x\cdot[\text{base}-1]$. Simplify the base expression and evaluate.
$L = \lim_{x\to\infty} x\left(\frac{e}{1-e}\cdot\frac{1}{x+1}\right) = \frac{e}{1-e}$. So $\log_e\alpha = \frac{e}{1-e}$. $\frac{\log_e\alpha}{1+\log_e\alpha} = \frac{e/(1-e)}{1+e/(1-e)} = \frac{e/(1-e)}{1/(1-e)} = e$.
Correct Answer: $e$