Wednesday, September 28, 2016

real analysis - T/F: a smooth function that grows faster than any linear function grows faster than x1+epsilon


Prove or find a counterexample to the claim that a smooth function that grows faster than any linear function grows faster than x1+ϵ for some ϵ>0.


My attempt: I understand that the first part of the problem claims lim. 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!


Answer




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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