Proving a limit of this growing function

Question

Let f be a growing(increasing) function:$$f : (0,+\infty)\rightarrow(0,+\infty)$$

Prove that, if $$\lim_{t\to\infty}\frac{f(2t)}{f(t)}=1$$

then $$\lim_{t\to\infty}\frac{f(a\cdot t)}{f(t)}=1 ,\forall a\in \mathbb{R},a\geq1$$

I thought about what kind of a function it could be, but it just seems too general to prove this way. I assume if it should be a polynomial function, it would have to be a constant,but then again there is no reason to believe it's a polynomial function. But if I can't make a general outlook of the function I'm not quite certain how to even begin solving this.$\\$

I should prove this without using derivations/integrals. Hints would be appreciated.

Thanks in advance!

Answer 1

First, show that:

$$\lim_{x\to \infty} \frac{f(2nx)}{f(x)} = 1$$

for each integer $n \ge 1$. One way to do this is by induction. Hint: note that

$$\frac{f(2(n+1)x)}{f(x)} = \frac{f(2(n+1) x)}{f(2nx)} \times \frac{f(2nx)}{f(x)}$$

now for $a \ge 1$, use the squeeze theorem:

$$1 \le \frac{f(at)}{f(t)} \le \frac{f(2(\lfloor a \rfloor +1)t)}{f(t)}$$

Answer 2

Using induction prove that $f(2^{n}t)/f(t)\to 1$ for all positive integers $n$ and then for a given $a\geq 1$ choose $n$ (this is possible because $2^{n}\to\infty$ as $n\to\infty$) such that $a<2^{n}$ and then use squeeze on $$1\leq\frac{f(at)}{f(t)}\leq\frac{f(2^{n}t)}{f(t)}$$