Types of Infinity — Question 9

Question 9

Prove that $$\lim_{x\to\infty}\frac{\ln x}{\sqrt{x}}=0.$$

Question 9 - Solution

We compare the growth rates of the logarithmic function and the power function.

Both $\ln x$ and $\sqrt{x}$ are positive and differentiable for $x>0$. Apply L’Hôpital’s Rule to the limit $$\lim_{x\to\infty}\frac{\ln x}{\sqrt{x}}.$$

Differentiate the numerator and denominator: $$\frac{d}{dx}(\ln x)=\frac{1}{x}, \qquad \frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}.$$

Thus, $$\lim_{x\to\infty}\frac{\ln x}{\sqrt{x}} = \lim_{x\to\infty}\frac{\frac{1}{x}}{\frac{1}{2\sqrt{x}}} = \lim_{x\to\infty}\frac{2\sqrt{x}}{x}.$$

Simplify: $$\frac{2\sqrt{x}}{x}=\frac{2}{\sqrt{x}}.$$

Since $$\lim_{x\to\infty}\frac{1}{\sqrt{x}}=0,$$ it follows that $$\lim_{x\to\infty}\frac{2}{\sqrt{x}}=0.$$

Therefore, $$\lim_{x\to\infty}\frac{\ln x}{\sqrt{x}}=0.$$

$$\boxed{0}$$