Answer by lhf for Proving $f(x)=\frac{x^2}{1+x}$ is uniformly continuous on...
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...
View ArticleAnswer by Michael Hardy for Proving $f(x)=\frac{x^2}{1+x}$ is uniformly...
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...
View ArticleAnswer by Pacciu for Proving $f(x)=\frac{x^2}{1+x}$ is uniformly continuous...
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...
View ArticleAnswer by Davide Giraudo for Proving $f(x)=\frac{x^2}{1+x}$ is uniformly...
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...
View ArticleAnswer by user17762 for Proving $f(x)=\frac{x^2}{1+x}$ is uniformly...
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...
View ArticleProving $f(x)=\frac{x^2}{1+x}$ 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...
View Article
