1. ## Another Unif. Cont.

Prove or disprove (by counterex.) $f(x) = xlog(\frac{1}{x})$ is unif. cont. on (0,1) and prove or disprove $f(x) = \frac{cosx}{mx+b}$ is unif. cont. on (0,1) for all nonzer0 $m,b \in \mathbb{R}$

So I tried getting both into Lipschitz but was not successful. So all I know to do is for the first one is say: Given $\epsilon > 0$ set $\delta = ?$. If $x,a \in (0,1)$, then $|xlog\frac{1}{x}+alog\frac{1}{a}|\le |xlog\frac{1}{x}| + |alog\frac{1}{a}| < 2$ (it's always going to be < 2 because our interval is (0,1)). Then I say if $(x,a) \in (0,1)$ and $|x-a| < \delta$ then $|f(x) - f(a)| = |xlog\frac{1}{x}-alog\frac{1}{a}|$ .... How do I know what I should set delta to be? Also I am totally lost starting here. Let me know if these is even remotely on the right track because I don't know for sure. Thank you so much.

2. Originally Posted by Math2010
Prove or disprove (by counterex.) $f(x) = xlog(\frac{1}{x})$ is unif. cont. on (0,1) and prove or disprove $f(x) = \frac{cosx}{mx+b}$ is unif. cont. on (0,1) for all nonzer0 $m,b \in \mathbb{R}$

So I tried getting both into Lipschitz but was not successful. So all I know to do is for the first one is say: Given $\epsilon > 0$ set $\delta = ?$. If $x,a \in (0,1)$, then $|xlog\frac{1}{x}+alog\frac{1}{a}|\le |xlog\frac{1}{x}| + |alog\frac{1}{a}| < 2$ (it's always going to be < 2 because our interval is (0,1)). Then I say if $(x,a) \in (0,1)$ and $|x-a| < \delta$ then $|f(x) - f(a)| = |xlog\frac{1}{x}-alog\frac{1}{a}|$ .... How do I know what I should set delta to be? Also I am totally lost starting here. Let me know if these is even remotely on the right track because I don't know for sure. Thank you so much.
The function $f(x) = x\log(1/x)$ is not Lipschitz on (0,1), because its derivative becomes infinite as $x\to0$. But $\textstyle\lim_{x\to0}f(x)$ does exist (it is 0), as is $\textstyle\lim_{x\to1}f(x)$. So f extends to a continuous function on the closed interval [0,1] and is therefore uniformly continuous.

3. Originally Posted by Math2010
Prove or disprove (by counterex.) $f(x) = xlog(\frac{1}{x})$ is unif. cont. on (0,1) and prove or disprove $f(x) = \frac{cosx}{mx+b}$ is unif. cont. on (0,1) for all nonzer0 $m,b \in \mathbb{R}$

So I tried getting both into Lipschitz but was not successful. So all I know to do is for the first one is say: Given $\epsilon > 0$ set $\delta = ?$. If $x,a \in (0,1)$, then $|xlog\frac{1}{x}+alog\frac{1}{a}|\le |xlog\frac{1}{x}| + |alog\frac{1}{a}| < 2$ (it's always going to be < 2 because our interval is (0,1)). Then I say if $(x,a) \in (0,1)$ and $|x-a| < \delta$ then $|f(x) - f(a)| = |xlog\frac{1}{x}-alog\frac{1}{a}|$ .... How do I know what I should set delta to be? Also I am totally lost starting here. Let me know if these is even remotely on the right track because I don't know for sure. Thank you so much.
The second one's kind of a pain in the ***. But, clearly both $f(x)=\frac{1}{mx+b},g(x)=\cos(x)$ are uniformly continuous on $I=(0,1)$ since $\left|f'(x)\right|\leqslant\left|\frac{m}{b^2}\rig ht|,\left|g(x)\right|\leqslant 1$ and so they are Lipschitz. And, there is a theorem that says that since $f,g$ are uniformly continuous and $I$ is bounded that $f\cdot g$ is uniformly continuous on $I$.

That's the least painful way I can see.