Limits, Continuity & Differentiability
Limit Involving Integral (L'Hôpital / Expansion)
nta_pyq_2025_apr
Grade 12
Question:
If $\lim_{t \to 0}\!\left(\int_0^1 (3x+5)^t\,dx\right)^{1/t} = \dfrac{\alpha}{5e} \cdot \left(\frac{8}{5}\right)^{2/3}$, then $\alpha$ is equal to ___
Step-by-Step Solution
Key Concept: Write the limit as $\exp\!\left(\lim_{t\to0}\frac{1}{t}\ln\!\int_0^1(3x+5)^t dx\right)$ and evaluate using L'Hôpital's rule, noting the integral becomes $\int_0^1 1\,dx=1$ at $t=0$.
Limit $= \exp\!\left(\int_0^1\ln(3x+5)\,dx\right) = \exp\!\left(\frac{\ln8^8-\ln5^5-3}{5}\right) = \frac{8^{8/5}}{5e} = \frac{64}{5e}\cdot\left(\frac{8}{5}\right)^{2/3}$... comparing gives $\alpha = 64$.
Correct Answer: 64