1. ## Limits Q

Use the inequality $\displaystyle 0< \int_{1}^{\sqrt{x}}\frac{dt}{t}< \sqrt{x}$ to show $\displaystyle \lim_{x\rightarrow \infty }(\frac{\ln x}{x})=0$

2. Originally Posted by vuze88
Use the inequality $\displaystyle 0< \int_{1}^{\sqrt{x}}\frac{dt}{t}< \sqrt{x}$ to show $\displaystyle \lim_{x\rightarrow \infty }(\frac{\ln x}{x})=0$
$\displaystyle 0 < \int_1^{\sqrt x } {\frac{{dt}}{t}} < \sqrt x \,\Leftrightarrow \, 0 < \ln \sqrt x < \sqrt x \, \Leftrightarrow$

$\displaystyle \Leftrightarrow 0 < \ln x < 2\sqrt x \Leftrightarrow 0 < \frac{{\ln x}} {x} < \frac{2} {{\sqrt x }} \Leftrightarrow$

$\displaystyle \Leftrightarrow 0 < \mathop {\lim }\limits_{x \to \infty } \frac{{\ln x}} {x} < \mathop {\lim }\limits_{x \to \infty } \frac{2} {{\sqrt x }} = 0 \Rightarrow \mathop {\lim }\limits_{x \to \infty } \frac{{\ln x}} {x} = 0.$

3. Originally Posted by vuze88
Use the inequality $\displaystyle 0< \int_{1}^{\sqrt{x}}\frac{dt}{t}< \sqrt{x}$ to show $\displaystyle \lim_{x\rightarrow \infty }(\frac{\ln x}{x})=0$
Evaluating the integral gives us $\displaystyle 0<\ln(\sqrt{x})<\sqrt{x}\implies 0<\frac{1}{2}\ln(x)<\sqrt{x}$.

Note that $\displaystyle \lim_{x\to\infty}\frac{\ln(x)}{x} = 2\lim_{x\to\infty}\frac{\frac{1}{2}\ln(x)}{x}$

The above equations imply that

$\displaystyle 0<\lim_{x\to\infty}\frac{\ln(x)}{x}<2\lim_{x\to\in fty}\frac{\sqrt{x}}{x}$

Now we know that $\displaystyle 2\lim_{x\to\infty}\frac{\sqrt{x}}{x}=2\lim_{x\to\i nfty}\frac{1}{\sqrt{x}}=0$, so we have successfully squeezed the limit:

$\displaystyle 0<\lim_{x\to\infty}\frac{\ln(x)}{x}<0$

So it equals $\displaystyle 0$. $\displaystyle \square$