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 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: 6,878 Thanks: 2240 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: 17,739 Thanks: 1361 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: 6,878 Thanks: 2240 Math Focus: Mainly analysis and algebra Yes.
 June 5th, 2017, 03:23 PM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,878 Thanks: 2240 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

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