$ \lim_{x \to \infty}f+f'=0$

Question

let $a,b$ real numbers such that $a<b$ and $f$,$g$ continuos on $[a,b]$ and differntiable on $]a,b[$

suppoose that: for $x \in ]a,b[$

$ |f'(x)|\leq g'(x)$

then $|f(b)-f(a)|\leq g(b)-g(a)$

$*$application :

let $k$ a differentiable function on $\mathbb R$

now suppose that $ \lim_{x \to \infty}k+k'=0$

let $h(x)=e^xk(x)$

let $c>0$

i have proved that :$\exists A>0$ such that for all $x>A$

$|h'(x)| \leq ce^x$

i have to prove that $ \lim_{x \to \infty}k(x)=0$

my idea is to fix $A<B<x$

so $|h(x)-h(B)|\leq a(e^x-e^B)$

then $|e^xk(x)-e^Bk(B)|\leq c(e^x-e^B)$

what i can do to find that $|k(x)|<c$

please help me with this question

Wait, $h$ is only defined on $[a,b].$ Or is that a different $f?$ — Thomas Andrews, Jul 15 '23 at 13:55
See https://math.stackexchange.com/q/649028/42969 or https://math.stackexchange.com/q/648652/42969 or https://math.stackexchange.com/q/840957/42969 — Martin R, Jul 15 '23 at 13:58

score 1 · Answer 1 · answered Jul 15 '23 at 14:58

1

You're almost done - you have $|k(x)-e^{B-x}k(B)| \le c(1-e^{B-x}) \le c$ whenever $x>B$ , and then you can simply take limits to get $\limsup_{x\to \infty} |k(x)| \le c$.

answered Jul 15 '23 at 14:58

raj

251

$ \lim_{x \to \infty}f+f'=0$

1 Answers1