Limits, Continuity & Differentiability
1^∞ Form — L'Hôpital Twice
nta_pyq_2026_jan
Grade 12
Question:
Let $f:\mathbb{R}\to(0,\infty)$ be a twice differentiable function such that $f(3)=18$, $f'(3)=0$ and $f''(3)=4$. Then $\displaystyle\lim_{x\to1}\left(\log_e\left(\dfrac{f(2+x)}{f(3)}\right)^{\frac{18}{(x-1)^2}}\right)$ is equal to:
Step-by-Step Solution
Key Concept: $1^\infty$ form. Exponent: $\tfrac{18}{(x-1)^2}\ln\tfrac{f(2+x)}{f(3)}$. As $x\to1$, $f(2+x) o f(3)$, so $\ln\tfrac{f(2+x)}{f(3)}\to0$. Apply L'Hôpital twice.
$\log_e T=2$. Answer $=2$.
Correct Answer: 3