August 19th, 2019, 10:22 PM  #1 
Senior Member Joined: Dec 2015 From: somewhere Posts: 642 Thanks: 91  Hard Limit
Evaluate the limit without L'hôpital’s rule. $\displaystyle \lim_{y\rightarrow \infty }\frac{\ln(yn)}{y^n } \; $, $\displaystyle n\in \mathbb{N}.$ Last edited by skipjack; August 20th, 2019 at 03:05 AM. 
August 20th, 2019, 02:35 AM  #2 
Senior Member Joined: Dec 2015 From: somewhere Posts: 642 Thanks: 91 
$\displaystyle L=\lim_{y\rightarrow \infty} \frac{\ln(y)}{y^n } +\lim_{y\rightarrow \infty}\frac{\ln(n)}{y^n }=\lim_{y\rightarrow \infty}\frac{\ln(y)}{y^n }=\lim_{y\rightarrow \infty }\frac{\ln(e^y)}{e^{yn} }=\lim_{y\rightarrow \infty}n^{1} \frac{y}{e^y}=\lim_{y\rightarrow \infty}(en)^{1} \frac{1+y}{e^y }=(en)^{1}L$ Now the equation is $\displaystyle L=(en)^{1} L$. Since $\displaystyle y^{1y} < \frac{y}{e^y } < \frac{y}{1+y} \; $ the limit converges . $\displaystyle L=(ne)^{1} L \: $ has solution L=0. Last edited by idontknow; August 20th, 2019 at 03:15 AM. 
August 21st, 2019, 07:27 AM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,683 Thanks: 2664 Math Focus: Mainly analysis and algebra 
Your $n^{1}$ isn't correct when it appears. The limit $\displaystyle \lim_{x \to \infty} \frac{\ln^a{x}}{x^b} = 0$ where $a > 0$ and $b > 0$ is a standard result that is often simply quoted. Whether you are allowed to do that or not depends on what you have been given (or proved) as results. \begin{alignat}{2} 0 &< \frac1t &&< t^{c1} &&& (c > 0,\, t> 1) \\ \int_1^x 0 \,\mathrm dt &< \int_1^x \frac1t \,\mathrm dt &&< \int_1^xt^{c1}\,\mathrm dt &&&(x > 1) \\ 0 &< \ln x &&< x^c  \frac1c &&< x^c \\ 0 &< \frac{\ln^a x}{x^b} &&< x^{acb} &&& (a > 0, \, b > 0) \quad \text{raising to the power $a$ and dividing by $x^b$}\\ \end{alignat} With $c=\frac{b}{2a} > 0$ we have $acb=\frac{b}2 < 0$ and so, the limit of the left and righthand expressions as $x \to \infty$ is zero and thus the limit $$\lim_{x \to \infty} \frac{\ln^a{x}}{x^b} = 0 \qquad (a > 0, \, b > 0)$$ Last edited by v8archie; August 21st, 2019 at 07:30 AM. 

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