Limits, Continuity & Differentiability
$e$-type Limits at Infinity
nta_pyq_2025_apr
Grade 12
Question:
$\lim_{x \to \infty} \dfrac{(2x^2 - 3x + 5)(3x-1)^{x/2}}{(3x^2 + 5x + 4)\sqrt{(3x+2)^x}}$ is equal to:
$\frac{2}{\sqrt{3e}}$
$\frac{2e}{\sqrt{3}}$
$\frac{2}{3\sqrt{e}}$
$\frac{2e}{3}$
Step-by-Step Solution
Key Concept: Factor out dominant terms: ratio of quadratics $\to 2/3$, and handle the exponential factor $(3x-1)^{x/2}/(3x+2)^{x/2}$ using the $e$-limit formula.
Polynomial ratio $\to 2/3$. $(3x-1)^{x/2}/(3x+2)^{x/2} = \left(1-\frac{3}{3x+2}\right)^{x/2} \to e^{-1/2}$. Limit $= \frac{2}{3}\cdot e^{-1/2} = \frac{2}{3\sqrt{e}}$.
Correct Answer: $\frac{2}{3\sqrt{e}}$