If $\lim_{x\to \infty}f(x)=L$ then $\forall k >0$, $\lim_{x\to \infty}f(kx)=L$

Question

If $\lim_{x\to \infty}f(x)=L$ then $\forall k >0$, $\lim_{x\to \infty}f(kx)=L$

So far I have: $\forall \epsilon>0$ $\exists A>0$ such that if $x>A$ then $|f(x)-L|<\epsilon$.

Now $|f(kx)-L|= |f(kx)-f(x)+f(x)-L|\leq|f(kx)-f(x)|+|f(x)-L|<|f(kx)-f(x)|+\epsilon$, but how do I bound $|f(kx)-f(x)|$ to some other constant?

Answer 1

Take $B={A\over k}$, $x>B$ implies that $kx>A$, this implies that $|f(xk)-L|<\epsilon$.

Answer 2

Let $ \epsilon >0 $ be given then there exists an $A$ such that for every $y$ if $y>A$, then $|f(y)-L|<\epsilon.$

Let $y = kx$ , and you are done.

Since it is a universal qualifier, you may simply change x to kx as well.

If it is true for every x, it is true for kx as well.