Is $\log (1 + {x^2})$ uniformly continuous on $[0,\infty)$? Is  $\log (1 + {x^2})$ uniformly continuous?
Here is my attempt:
Let $\forall\left| {x - y} \right| < \delta$:
$\left| {\log (1 + {x^2}) - \log (1 + {y^2})} \right| = \left| {\log (\frac{{1 + {x^2}}}{{1 + {y^2}}})} \right| < \varepsilon $
because $\log (1) = 0$ it's sufficent to prove:  
$\left| {\frac{{1 + {x^2}}}{{1 + {y^2}}} - 1} \right| < \varepsilon $
This is where I got stuck. What do you suggest? 
EDIT:
I'd like to prove it without involving derivative. Namely, by the definition of uniform continuity. 
 A: The function $f:t\mapsto\log(1+t^2)$ is differentiable hence, by the mean value theorem, for every $x\lt y$, there exists some $z$ in $(x,y)$ such that $f(y)-f(x)=f'(z)(y-x)$. Since $f'(z)=2z/(1+z^2)$, $|f'(z)|\leqslant1$ hence $|f(y)-f(x)|\leqslant|y-x|$ for every $x$ and $y$. The function $f$ is $1$-Lipschitz, in particular, $f$ is uniformly continuous (for every $\varepsilon$, choose $\delta=\varepsilon$).
A: Let me give you a more direct method (that is, one that relies on less machinery).
Let $\Delta y:=x-y$, so that $x=y+\Delta y$. Then
$$
\frac{1+x^2}{1+y^2}=\frac{1+y^2+2y\Delta y+(\Delta y)^2}{1+y^2}=1+\frac{2y\Delta y}{1+y^2}+\frac{(\Delta y)^2}{1+y^2}.
$$
From here, you can note that for all $y\in\mathbb{R}$,
$$
\left\lvert\frac{2y}{1+y^2}\right\rvert\leq 1\qquad\text{and}\qquad\left\lvert\frac{1}{1+y^2}\right\rvert\leq 1.
$$
(The second inequality is immediate; for the first try considering $\lvert y\rvert\leq 1$ and $\lvert y\rvert>1$ as separate cases.)
But, as a result,
$$
\left\lvert\frac{1+x^2}{1+y^2}-1\right\rvert\leq\left\lvert\frac{2y}{1+y^2}\right\rvert\lvert\Delta y\rvert+\left\lvert\frac{1}{1+y^2}\right\rvert(\Delta y)^2\leq\lvert\Delta y\rvert+(\Delta y)^2.
$$
You can definitely make this last expression small by making $\Delta y$ small... and note that $\lvert\Delta y\rvert<\delta$.
A: Hint: Yes. You can use the mean value theorem. See a related problem.
