L’Hospital’s Rule and Indeterminate Forms — Question 5

Question 5 - Solution

As $x \to \infty$, both numerator and denominator approach infinity: $$\frac{\ln x}{\sqrt{x}} \to \frac{\infty}{\infty} \quad \text{(indeterminate form)}$$

Apply L’Hospital’s Rule: $$\lim_{x \to \infty} \frac{\ln x}{\sqrt{x}} = \lim_{x \to \infty} \frac{\frac{d}{dx} \ln x}{\frac{d}{dx} \sqrt{x}} = \lim_{x \to \infty} \frac{\frac{1}{x}}{\frac{1}{2\sqrt{x}}} = \lim_{x \to \infty} \frac{2\sqrt{x}}{x} = \lim_{x \to \infty} \frac{2}{\sqrt{x}} = 0$$

Final Answer: $$\boxed{0}$$