Proving $f(x)=\frac{x^2}{1+x}$ is uniformly continuous on $[0,\infty)$ I have a homework assignment to complete and I am having trouble proving that $f(x)=\frac{x^2}{1+x}$ is uniformly continuous on $[0,\infty)$.
For some reason I can't find the way to solve this one...
Cans someone please help me? Thanks :) 
EDIT: I made a mistake.  I meant to write  $$f(x)=\frac{x^2}{1+x}$$ rather than $f(x) = \frac{x}{1+x}$.
 A: We can write, since $x,y\geq 0$ 
$$|f(x)-f(y)|=\left|1-\frac 1{1+x}-\left(1-\frac 1{1+y}\right)\right|=\frac{|1+y-(1-x)|}{(1+x)(1+y)}=\frac{|x-y|}{(1+x)(1+y)}\leq |x-y|$$
so $f$ is $1$-Lipschitz, and uniformly continuous (take $\delta:=\varepsilon$ in the definition).
For $f(x)=\frac{x^2}{1+x}$, write 
$$f(x)=x\frac{x+1-1}{1+x}=x-\frac x{1+x}=x-\frac{x+1-1}{1+x}=x-1+\frac 1{1+x}, $$
so 
$$|f(x)-f(y)|=\left|x-1+\frac 1{1+x}-\left(y-1+\frac 1{1+y}\right)\right|\leq 2|x-y|$$
(so this time you have to take $\delta:=\frac{\varepsilon}2$).
A: Any function which has bounded derivative on the underlying set is uniformly continuous on the underlying set. More generally, any function which is Hölder continuous on the underlying set is uniformly continuous on the underlying set. In your example, where $$f(x) = \frac{x}{1+x} = 1 - \frac1{1+x}$$ we have that $$f'(x) = \frac1{(1+x)^2}$$
On the set $[0, \infty)$, we have that $$\lvert f'(x)\rvert \leq 1.$$
Hence, $\displaystyle f(x) = \frac{x}{1+x}$ is uniformly continuous on $[0,\infty)$.
A: Let $f(x):=\frac{x^2}{1+x}$ for $x\geq 0$.
Function $f$ is continuous in $[0,\infty[$ and its graph has an oblique asymptote when $x$ tends to $\infty$ (the asymptote being the straight line of equation $y=x-1$).
Finally you can conclude using the following lemma:

Let $a\in \mathbb{R}$ and $f:[a,\infty[\to \mathbb{R}$ a continuous function.
If the graph of $f$ has a horizontal or an oblique asymptote when $x$ tends to $\infty$, then $f$ is uniformly continuous in $[a,\infty[$.


The lemma cited above is a corollary of the following theorem:

Let $a\in \mathbb{R}$ and $f,g:[a,\infty[\to \mathbb{R}$ be continuous functions such that:
  $$\tag{A} \lim_{x\to \infty} f(x)-g(x)=0\; .$$
  Then $f$ is uniformly continuous in $[a,\infty[$ if and only if $g$ is uniformly continuous in $[a,\infty[$.

The proof is very simple.
Since the statement is symmetric w.r.t. $f$ and $g$, it suffices to prove only one implication; therefore we assume, e.g., that $g$ is u.c. in $[a,\infty[$ and prove that also $f$ is u.c. in $[a,\infty[$. Moreover, w.l.o.g., we assume also $a=0$.
Let $\varepsilon >0$ be fixed. By (A) we can find $r\geq 0$ such that:
$$\tag{I} \forall x\geq r,\quad |f(x)-g(x)|\leq \frac{\varepsilon}{8}\; ;$$
since $g$ is u.c. in $[0,\infty[$ we can find $\delta_1>0$ such that:
$$\tag{II} \forall x,y\geq 0,\quad |x-y|\leq \delta_1\ \Rightarrow\ |g(x)-g(y)|\leq \frac{\varepsilon}{4}\; ;$$
finally, by Heine-Cantor theorem, we can find $\delta_2>0$ such that:
$$\tag{III} \forall 0\leq x,y\leq r,\quad |x-y|\leq \delta_2\ \Rightarrow\ |f(x)-f(y)|\leq \frac{\varepsilon}{2}\; .$$
Now, set $\delta:=\min \{\delta_1, \delta_2\}$ and take two arbitrary $x,y\geq 0$ s.t. $|x-y|\leq \delta$: then, due to the symmetry w.r.t. $x,y$, three cases arise:


*

*$r\leq x,y$: in this case we find:
$$|f(x)-f(y)|\leq |f(x)-g(x)|+|g(x)-g(y)|+|g(y)-f(y)|\; ,$$
hence $|f(x)-f(y)|\leq \frac{\varepsilon}{8}+\frac{\varepsilon}{4}+\frac{\varepsilon}{8}=\frac{\varepsilon}{2}\leq \varepsilon$ because of (I) & (II)

*$0\leq x,y\leq r$: in such a case $|f(x)-f(y)|\leq \frac{\varepsilon}{2}\leq \varepsilon$ by (III);

*$0\leq x\leq r < y$: in this case we have:
$$|f(x)-f(y)|\leq |f(x)-f(r)| +|f(r)-f(y)|\; ,$$
thus $|f(x)-f(y)|\leq \frac{\varepsilon}{2} +\frac{\varepsilon}{2}=\varepsilon$ for (III) & case 1.
Therefore $|x-y|\leq \delta\ \Rightarrow\ |f(x)-f(y)|\leq \varepsilon$ for all $x,y\geq 0$ and $f$ is u.c. in $[0,\infty[$. $\square$
A: On $[0,\infty)$ the derivative of this function is non-negative but always less than $1$, so $|f(u)-f(v)| < |u-v|$ for all $u,v$ (you can use the mean value theorem to justify what comes after the word "so" above).  Consequently you can just take $\delta=\varepsilon$.
A: As others have noticed, $$f(x)=x-1+\frac 1{1+x}$$
Now this is a sum of the identity function, which certainly is uniformly continuous, with a bounded decreasing function (one that even goes to $0$ as $x\to\infty$), which also is uniformly continuous. Hence $f$ is uniformly continuous.
