Is $f(x)=x+\frac{x}{x+1}$ uniformly continuous on $(0,\infty)$

Question

Is $f(x)=x+\frac{x}{x+1}$ uniformly continuous on $(0,\infty)$

Going from the epsilon delta definition we get:

$$\forall x,y>1,\text{WLOG}:x>y \ ,\ \forall\epsilon>0,\exists\delta>0 \ s.t. \ |x-y|<\delta\rightarrow |x+\frac{x}{x+1}-y-\frac{y}{y+1}|<\epsilon$$

But I'm not really sure on how to continue from here.

Alternatively: if $\lim\limits_{x\to \infty}f(x)$ exists on $[0,\infty)$ then the function is uniformly continuous, well, it's easy to see here that $\lim\limits_{x\to \infty}f(x)=\infty$ but is that enough ?

Well, is it enough that $\lim_{x\to\infty}f(x)=\infty$? You should be able to answer that. — Carsten S, Dec 25 '13 at 20:38

score 3 · Accepted Answer · edited Dec 25 '13 at 20:19

Answer here if you want it, I hope that the spoiler tags work. (If not, could someone tell me how to spoiler multiple lines?)

Let $a \in (0,\infty)$ be given.

Let $\epsilon >0$ be given.

Select $\delta=\frac{\epsilon}{2}$

Then, for all $x \in (0,\infty)$ with $|x-a|<\delta$:

$$|f(x) - f(a)|=\left|x+\frac{x}{x+1} - a -\frac{a}{a+1}\right| \leq |x-a| + \left|\frac{x-a}{(x+1)(a+1)}\right|$$

$$=|x-a|+\frac{|x-a|}{|x+1||a+1|}\leq |x-a| + |x-a| = 2|x-a| < 2\delta \leq 2\cdot \frac{\epsilon}{2}=\epsilon$$

Hence, $f$ is continuous at $a$. Since $a \in (0,\infty)$ and $\epsilon >0$ are given arbitrarily, we have: for all $ a \in (0,\infty)$ and all $ \epsilon > 0$, there exists a $\delta > 0$ such that for all $x \in (0,\infty)$ with $|x-a|<\delta$, $|f(x) - f(a)|<\epsilon$. Conclusion: $f$ is continuous.

To do multiple line spoiler tags, put >! like you did, but remove all those extra line breaks. Where you want line breaks, use <br><br> — Justin, Dec 25 '13 at 20:18

score 2 · Answer 2 · answered Dec 25 '13 at 17:56

2

Take a smaller bite. Note that a sum of uniformly continuous functions is uniformly continuous. We know $x\mapsto x$ is uniformly continuous, so we just deal with $x/(x+1)$. Also, notice that $${x\over x + 1 } = 1 - {1\over x + 1}.$$ Now the job is to show that $x\mapsto 1/(x + 1)$ is uniformly continuous. Can you do the rest?

answered Dec 25 '13 at 17:56

ncmathsadist

49,383

So you're saying that I do have to find an epsilon, but modifying the function should help. – GinKin Dec 25 '13 at 18:43
With this function, it is easy to do the $\epsilon$-$\delta$ tango. – ncmathsadist Dec 25 '13 at 18:47
@GinKin, that limit criterion might come in handy for some of the summands. – Carsten S Dec 25 '13 at 20:39

score 1 · Answer 3 · edited Apr 13 '17 at 12:21

1

Hint: Use the mean value theorem. Note that, the derivative of the function is bounded on $(0,\infty)$. See a related problem.

edited Apr 13 '17 at 12:21

Community

1

answered Dec 25 '13 at 17:54

Mhenni Benghorbal

47,431

Thanks. In one of the answers there they use zeta which I have no idea what it is so we haven't covered that. – GinKin Dec 25 '13 at 18:40
1

I think using the mean value theorem is the easiest to prove U.continuity since $|f'(x)| \leq 2 $ on $(0,infty)$. – Mhenni Benghorbal Dec 25 '13 at 18:43
@Downvoter: What's the downvote for? – Mhenni Benghorbal Dec 25 '13 at 20:22
3 out of 4 answers here got downvoted. – GinKin Dec 25 '13 at 20:54
@GinKin: Do not worry about the down vote. – Mhenni Benghorbal Dec 25 '13 at 20:56

Is $f(x)=x+\frac{x}{x+1}$ uniformly continuous on $(0,\infty)$

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