September 21st, 2015, 08:19 AM  #1 
Senior Member Joined: Dec 2014 From: The Asymptote Posts: 142 Thanks: 6 Math Focus: Certainty  Limit proofs
I have to limits which I wish to prove 1. $\displaystyle \lim_{x\to \infty} \frac{e^x}{x^n} = \infty$. Prove that for any positive integer n. The exponential function approaches infinity faster than any power of x. 2. $\displaystyle \lim_{x\to \infty} \frac{lnx}{x^p} = 0$. Prove that for any number p > 0. The logarithmic function approaches infinity more slowly than any power of x. How to do so mathematically without the need to plug in values??? 
September 21st, 2015, 01:49 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,451 Thanks: 567 
For 1, L'Hopital's rule ends up with $\displaystyle lim_{x\to \infty }\frac{e^x}{n!}$. For 2, L'Hopital's rule gives $\displaystyle lim_{x\to \infty }\frac {\frac{1}{x}}{px^{p1}}=\frac{1}{px^p}$. 
September 21st, 2015, 02:14 PM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,229 Thanks: 2411 Math Focus: Mainly analysis and algebra 
If you write $\displaystyle 0 \lt \frac1t \lt t^{c1}$ for $\displaystyle t \gt 1$ and $\displaystyle c \gt 0$ you can integrate with respect to $\displaystyle t$ between $\displaystyle 1$ and $\displaystyle x \gt 1$. Then raise each part of the inequality to the power $\displaystyle a \gt 0$ and divide by $\displaystyle x^b$ where $\displaystyle b \gt 0$. Finally, choose $\displaystyle c = {b \over 2a} \gt 0$ and use the squeeze theorem to prove a more general result than the second limit: $\displaystyle \lim_{x \to \infty} {\log^a x \over x^b} = 0$Setting $\displaystyle x = \mathrm e^y$ then gives a more general result than the first limit. 

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