# Proving a limit of this growing function

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.

• I'm the \langle \rangle fairy, here to let you know that $\langle, \rangle$ plays nicer with TeX than <, > does :) – Patrick Stevens Jan 28 '17 at 13:56
• @PatrickStevens I see :) I'm still relatively new to Tex, so I may still have some mistakes, thanks for your help, I will edit my answer! – Collapse Jan 28 '17 at 14:06

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:

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

• Thanks a lot, awesome answer!! – Collapse Jan 28 '17 at 14:06
• Your induction step is wrong. – Paramanand Singh Jan 29 '17 at 8:46
• @ParamanandSingh please be specific. – user384138 Jan 29 '17 at 9:57
• You have proved that $f(2nx)/f(x)\to 1$ whereas it should be $f(2^{n}x)/f(x)\to 1$. Specifically you can't say anything about the limit $f(2(n+1)x)/f(2nx)$ from the conditions given in question. That's the flaw in your reasoning. And it is strange that people don't notice this before up voting. – Paramanand Singh Jan 29 '17 at 10:44
• @ParamanandSingh you're right. Thank you. – user384138 Jan 29 '17 at 10:49

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 $$1\leq\frac{f(at)}{f(t)}\leq\frac{f(2^{n}t)}{f(t)}$$