Asymptotes and uniform continuity

Let $f:\mathbb{R}\to \mathbb{R}$ be a continuous function with oblique asymptotes for $x\to \pm \mathrm{\infty}$. Prove that the function is uniformly continuous.

Proof: Let . Since the function has obliquous asymptotes then

$\mathrm{\exists}M>0:|f(x)-(ax+b)|<\u03f5\phantom{\rule{1em}{0ex}}\mathrm{\forall}\phantom{\rule{thickmathspace}{0ex}}x>M$

and

$\mathrm{\exists}m>0:|f(x)-(cx+d)|<\u03f5\phantom{\rule{1em}{0ex}}\mathrm{\forall}\phantom{\rule{thickmathspace}{0ex}}x<-m$

Now, in the interval the function is uniformly continuous because of the Heine-Cantor theorem. We deal with the case $+\mathrm{\infty}$ since the other one is formally identical. We know that lines are uniformly continuous. So, using the that certainly exists for the line we have:

$\begin{array}{rl}|f(x)-f({x}_{0})|& =|f(x)-(a{x}_{0}+b)+(a{x}_{0}+b)-f({x}_{0})|\\ & \le |f(x)-(a{x}_{0}+b)|+|(a{x}_{0}+b)-f({x}_{0})|\\ & <|f(x)-(a{x}_{0}+b)|+\u03f5\end{array}$

But

$|f(x)-(a{x}_{0}+b)|\le |f(x)-(ax+b)|+|(ax+b)-(a{x}_{0}+b)|<2\u03f5$

So we are finished because if a continuous function is uniformly continuous in and , then it is uniformly continuous in .

Let $f:\mathbb{R}\to \mathbb{R}$ be a continuous function with oblique asymptotes for $x\to \pm \mathrm{\infty}$. Prove that the function is uniformly continuous.

Proof: Let . Since the function has obliquous asymptotes then

$\mathrm{\exists}M>0:|f(x)-(ax+b)|<\u03f5\phantom{\rule{1em}{0ex}}\mathrm{\forall}\phantom{\rule{thickmathspace}{0ex}}x>M$

and

$\mathrm{\exists}m>0:|f(x)-(cx+d)|<\u03f5\phantom{\rule{1em}{0ex}}\mathrm{\forall}\phantom{\rule{thickmathspace}{0ex}}x<-m$

Now, in the interval the function is uniformly continuous because of the Heine-Cantor theorem. We deal with the case $+\mathrm{\infty}$ since the other one is formally identical. We know that lines are uniformly continuous. So, using the that certainly exists for the line we have:

$\begin{array}{rl}|f(x)-f({x}_{0})|& =|f(x)-(a{x}_{0}+b)+(a{x}_{0}+b)-f({x}_{0})|\\ & \le |f(x)-(a{x}_{0}+b)|+|(a{x}_{0}+b)-f({x}_{0})|\\ & <|f(x)-(a{x}_{0}+b)|+\u03f5\end{array}$

But

$|f(x)-(a{x}_{0}+b)|\le |f(x)-(ax+b)|+|(ax+b)-(a{x}_{0}+b)|<2\u03f5$

So we are finished because if a continuous function is uniformly continuous in and , then it is uniformly continuous in .