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 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 $\displaystyle y^{1-y} < \frac{y}{e^y } < \frac{y}{1+y} \;$ 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^{c-1} &&& (c > 0,\, t> 1) \\ \int_1^x 0 \,\mathrm dt &< \int_1^x \frac1t \,\mathrm dt &&< \int_1^xt^{c-1}\,\mathrm dt &&&(x > 1) \\ 0 &< \ln x &&< x^c - \frac1c &&< x^c \\ 0 &< \frac{\ln^a x}{x^b} &&< x^{ac-b} &&& (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 $ac-b=-\frac{b}2 < 0$ and so, the limit of the left- and right-hand expressions as $x \to \infty$ is zero and thus the limit $$\lim_{x \to \infty} \frac{\ln^a{x}}{x^b} = 0 \qquad (a > 0, \, b > 0)$$ Thanks from topsquark Last edited by v8archie; August 21st, 2019 at 07:30 AM. Tags hard, limit Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post idontknow Real Analysis 3 June 15th, 2019 05:34 PM Luciferis Calculus 22 December 14th, 2016 02:44 AM kmaira8 Calculus 5 June 16th, 2014 02:37 PM stuart clark Calculus 1 March 14th, 2011 09:33 PM 4011 Calculus 12 May 7th, 2007 04:23 PM

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