<s>[INST] $\sqrt{x^2+1}$ uniformly continuous on (0, 1)? $\sqrt{x^2+1}$ uniformly continuous on (0, 1)?
How to deal with such problems? Please help.
 I know the definition of U C. but unable to handle the problem. 
 [/INST] Hint: Observe
\begin{align}
\left|\sqrt{x^2+1}-\sqrt{y^2+1}\right| =& \frac{|x^2-y^2|}{\sqrt{x^2+1}+\sqrt{y^2+1}} \\
=&\  \frac{|x+y|}{\sqrt{x^2+1}+\sqrt{y^2+1}}|x-y|\\
 \leq&\ \frac{|x|+|y|}{2}|x-y| \leq |x-y|
\end{align}
when $x, y \in (0, 1)$.
Edit: This is more or less the solution. However, you should try to use the definition of uniform continuity to fill in the details.
More Edit: To show $f(x)= \sqrt{x^2+1}$ is uniformly continuous on all of $\mathbb{R}$ observe
\begin{align}
\frac{|x|+|y|}{\sqrt{x^2+1}+\sqrt{y^2+1}} \leq \frac{|x|}{\sqrt{x^2+1}}+\frac{|y|}{\sqrt{y^2+1}}\leq 2.
\end{align}
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