Let f(x) be continous on the real number line. Suppose that lim(x-> infinity) f(x) = 0 and lim(x-> negative infinity) f(x) = 0. Show that f(x) is unformly continous or else give a counterexample.
Let f(x) be continous on the real number line. Suppose that lim(x-> infinity) f(x) = 0 and lim(x-> negative infinity) f(x) = 0. Show that f(x) is unformly continous or else give a counterexample.
More generally, if $\displaystyle \lim_{x\to\infty}f(x)=L_1,\lim_{x\to-\infty}f(x)=L_2<\infty$ and $\displaystyle f$ is continuous then it's uniformly continuous.
Start like this, for some $\displaystyle \varepsilon>0$ choose $\displaystyle A>0$ such that $\displaystyle |f(x)-L_1|<\frac{\varepsilon}{2},\text{ }x\in[A,\infty)$ AND $\displaystyle |f(x)-L_2|<\frac{\varepsilon}{2},\text{ }x\in(-\infty,-A]$.
You clearly have then that $\displaystyle |f(x_1)-f(x_2)|<\varepsilon,x_1,x_2\in\mathbb{R}-[-A,A]$ and you surely have that $\displaystyle f$ is unif. cont. on $\displaystyle [-A,A]$. See if you piece the riece together.
Why does it follow that it is uniformly continuous on the interval [-M,M]? I can see the intervals (-inf,-M] and [M,inf) , since the delta neighborhood of x values need not depend on x, only epsilon as the slope varys less and less towards the ends of the domain.