User Name Remember Me? Password

 Pre-Calculus Pre-Calculus Math Forum

 June 5th, 2017, 09:00 AM #1 Newbie   Joined: Jun 2017 From: Bucharest Posts: 5 Thanks: 0 Impossible limit! Sorry for the title. There is a limit that's giving me hell. I cannot solve it, but I know the answer is 0. lim x->0 with x<0 from f(x)=(e^(1/x))/(x^2) I sure know the limit is 0, but I don't know why. It is a vertical asymptote and for x>0 the limit is +inf but for x<0, I don't know how to calculate it. Last edited by skipjack; June 5th, 2017 at 11:12 AM. June 5th, 2017, 11:11 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,689 Thanks: 2669 Math Focus: Mainly analysis and algebra Write $y=-\frac1x$ so that $$\lim_{x \to 0^-} \frac{e^\frac1x}{x^2}=\lim_{y \to +\infty} \frac{y^2}{e^y}$$ There is a standard result that $$\lim_{t \to +\infty} \frac{t^a}{e^{bt}}=0 \quad \text{for all positive a and b}$$ Thanks from Alexxandru June 5th, 2017, 11:17 AM #3 Global Moderator   Joined: Dec 2006 Posts: 21,015 Thanks: 2250 Let x = -1/u, then it's the limit of u²/e^u as u tends to positive infinity. How you prove that limit is zero depends on how much you know about the exponential function. Thanks from Alexxandru June 5th, 2017, 01:43 PM #4 Newbie   Joined: Jun 2017 From: Bucharest Posts: 5 Thanks: 0 Thanks very much guys. So basically the exponential function grows faster than polynomial so that's why it tends to 0. Last edited by skipjack; June 5th, 2017 at 09:41 PM. June 5th, 2017, 02:24 PM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,689 Thanks: 2669 Math Focus: Mainly analysis and algebra Yes. June 5th, 2017, 03:23 PM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,689 Thanks: 2669 Math Focus: Mainly analysis and algebra For $c \gt 0$ and $t \gt 1$ we have $$\begin{array}{c l} 0 \lt \tfrac1t \lt t^{c-1} & \text{and for x \gt 1} \\ 0 \lt \int_1^x \tfrac1t \mathrm dt \lt \int_1^x t^{c-1} \mathrm dt \\ 0 \lt \ln{x} \lt \tfrac1c x^c - 1 \lt \tfrac1c x^c \\ \\ 0 \lt \ln^a{x} \lt \tfrac1c x^{ac} & (a \gt 0) \\ 0 \lt \frac{\ln^a{x}}{x^b} \lt \tfrac1c x^{ac-b} & (b \gt 0) \\ & \text{now, pick c = \frac{b}{2a} \gt 0} \\ 0 \lt \frac{\ln^a{x}}{x^b} \lt \tfrac1c x^{-\frac{b}{2}} \\ \end{array}$$ and since $-\frac{b}{2} \lt 0$ the squeeze theorem gives $$\lim_{x \to \infty} \frac{\ln^a{x}}{x^b} = 0$$ Now with $y = \ln x$ we get $$\lim_{y \to \infty} \frac{y^a}{e^{by}} = 0$$ for any positive $a$ and $b$. Thanks from Alexxandru Tags impossible, 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 Kevineamon Linear Algebra 7 March 10th, 2017 07:52 AM Samuel Leite Geometry 1 September 23rd, 2016 03:54 PM Ole Daniel Algebra 3 January 13th, 2015 12:50 AM Etyucan Calculus 2 October 25th, 2011 08:22 PM ITConsultant1996 Elementary Math 3 June 24th, 2009 11:50 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top      