Integration inequality question help: Sketch the curve y=1/u for u > 0... Sketch the curve $y=\frac{1}{u}$ for $u > 0$. From the diagram, show that $\int\limits_1^{\sqrt{x}}\frac{du}{u}< \sqrt x-1$, for x > 1. Use this result to show that $0 < \ln(x) < 2(\sqrt{x}-1)$ and hence that $\frac{\ln x}{x}\to0$ as $x\to\infty$. Deduce that $x\ln(x)\to0$ as $x\to0$.
Basically, I figured out the first part in terms of it being from the diagram. From 'use this result' in the question, I don't know what to do next. Can anyone help me out? 
Thank you!

Big Hint: 
We know by the fundamental theorem of calculus that 
$$\int_{1}^{\sqrt{x}} \frac{du}{u} = \frac{ln(x)}{2}.$$ 
However,
$$\int_{1}^{\sqrt{x}} \frac{du}{u} < \sqrt{x} -1.$$
Thus, we can say $$0<\frac{ln(x)}{2} < \sqrt{x} -1 \text{ or } 0<lnx<2(\sqrt{x} -1).$$ 
Now, if we divide by $x$ and apply Squeeze Theorem, one will quickly find what they asked for. 
