Is $f(x) = \log(1+x^2)$ uniformly continuous on $(0,\infty)$?
My work:
Looking at the graph and knowing that $\log$ considered a "slow-growing" function, my guess is that $f(x)$ is uniformly continuous.
Let $\varepsilon > 0$ and $\left| {x - y} \right| < \delta $ where $\delta$ will be defined later.
$$\left| {f(x) - f(y)} \right| = \left| {\log (1 + {x^2}) - \log (1 + {y^2})} \right| = \left| {\log \left( {{{1 + {x^2}} \over {1 + {y^2}}}} \right)} \right| \le {{1 + {x^2}} \over {1 + {y^2}}}$$
The problem is that the last expression is too large.
I tried assuming $x < y$ (without loss of generality).
Also, I tried multiplying by $(1-y^2)$ but nothing good came out from that.
What am I missing?
Update:
I'm not allowed to use the fact that the derivative is bounded which implies uniform continuity. I'm suppose to prove it "directly".
Hint: $f\in\mathcal{C}^1$ (i.e., is continuously differentiable), and its derivative is bounded. Hence, it is Lipschitz.
Another approach (following the edit of the original question -- "Lipschitz-ness is not allowed"): suppose wlog that $\epsilon \leq \frac{1}{2}$, and let $\delta\stackrel{\rm{def}}{=}\frac{\epsilon}{2}$. Fix any $x \geq y$ with $\lvert x-y \rvert \leq \delta$. Then $$ 1 \leq \frac{1+x^2}{1+y^2} \leq \frac{1+(y+\delta)^2}{1+y^2} = 1+\delta \frac{2y+\delta}{1+y^2} \leq 1+\delta \frac{1+2y}{1+y^2} $$ so that $$ \ln 1 \leq \ln \frac{1+x^2}{1+y^2} \leq \ln\!\left( 1+\delta \frac{1+2y}{1+y^2} \right) $$ Now, using the "fact" (easy to prove) that for every $y > 0$ one has $\frac{1+2y}{1+y^2} \leq \frac{1+\sqrt{5}}{2} \leq 2$, you get $$ 0 \leq \ln \frac{1+x^2}{1+y^2} \leq \ln(1+2\delta)\leq 2\delta = \epsilon. $$