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Prove or find a counterexample to the claim that a smooth function that grows faster than any linear function grows faster than $x^{1+\epsilon}$ for some $\epsilon>0$.

My attempt: I understand that the first part of the problem claims $\lim_{x\rightarrow \infty}\frac{g(x)}{kx} = \infty, \forall k>0$. We want to show, then, that $\exists \epsilon >0$ and constant $l>0$ such that $\lim_{x\rightarrow \infty}\frac{g(x)}{lx^{1+\epsilon}} = \infty$.

I've tried using the definition of limits, but I get stuck trying to bound the function $\frac{1}{x^\epsilon}$. Also, I've tried using L'Hopital's rule to no avail. Any ideas?

Any help is appreciated!

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  • $\begingroup$ Try $g(x) = x \ln x$. $\endgroup$ Commented Jan 27, 2017 at 19:15
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    $\begingroup$ The way to approach this is to find a function that does not grow as fast as $x^{\varepsilon}$ for any $\varepsilon$ then just multiply it by $x$, as per @OpenBall's answer. $\endgroup$ Commented Jan 27, 2017 at 19:19
  • $\begingroup$ After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark ✓ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?. $\endgroup$ Commented Jan 29, 2017 at 17:07

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Hint: It is false. Find a counterexample.

Followup hint: (place your mouse on the hidden text to show it)

The function $f\colon(0,\infty)\to\mathbb{R}$ defined by $f(x) = x\ln x$ is such a counterexample.

Followup followup hint: (pretty much the solution, with some details to fill in; place your mouse on the hidden text to show it)

For any $a>0$, $\frac{x\ln x}{a x} = \frac{1}{a}\ln x \xrightarrow[x\to\infty]{} \infty$. However, for any fixed $\epsilon > 0$, $$\frac{x\ln x}{x^{1+\epsilon}} = \frac{\ln x}{x^\epsilon}=\frac{1}{\epsilon}\frac{\ln(x^\epsilon)}{x^\epsilon} = \frac{1}{\epsilon}\frac{\ln t}{t}$$ for $t=x^\epsilon \xrightarrow[x\to\infty]{}\infty$.

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