Prove that $\lim_{x\to\infty} \frac{f(x)}{x^{2}}=0$ for uniformly continuous function $f$

Asked Jul 04 '18 at 14:58

Active Jul 04 '18 at 14:58

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Prove that $\lim_{x\to\infty} \frac{f(x)}{x^{2}}=0$ for uniformly continuous $f:[1,\infty)\to\mathbb{R}$.

I don't actually know if this is true, but I guess it should be, because functions with too strong (more than ~$x$) growth at infinity shouldn't be uniformly continuous (which is small changes in the scale of ~$x$). Specifically, I can show that for a monomic $f(x)=x^k$, it is uniformly continuous if and only if $k\geq1$, which then satisfy the desired limit.

I would also use a Taylor expansion had I known $f$ is differentiable.

asked Jul 04 '18 at 14:58

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Do you mean $k\leq 1$? – Clement C. Jul 04 '18 at 15:03
And your guess is correct: see e.g. Theorem 3.1. – Clement C. Jul 04 '18 at 15:05
See https://math.stackexchange.com/questions/305014/every-uniformly-continuous-real-function-has-at-most-linear-growth-at-infinity – Robert Z Jul 04 '18 at 15:05
Also this: https://math.stackexchange.com/q/539998/144766 – mechanodroid Jul 04 '18 at 15:20

Prove that $\lim_{x\to\infty} \frac{f(x)}{x^{2}}=0$ for uniformly continuous function $f$

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