June 5th, 2017, 10:00 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 12:12 PM. 
June 5th, 2017, 12:11 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,094 Thanks: 2360 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 $$\lim_{t \to +\infty} \frac{t^a}{e^{bt}}=0 \quad \text{for all positive $a$ and $b$}$$ 
June 5th, 2017, 12:17 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 18,250 Thanks: 1439 
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.

June 5th, 2017, 02:43 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 10:41 PM. 
June 5th, 2017, 03:24 PM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,094 Thanks: 2360 Math Focus: Mainly analysis and algebra 
Yes.

June 5th, 2017, 04:23 PM  #6 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,094 Thanks: 2360 Math Focus: Mainly analysis and algebra 
For $c \gt 0$ and $t \gt 1$ we have $$\begin{array}{c l} 0 \lt \tfrac1t \lt t^{c1} & \text{and for $x \gt 1$} \\ 0 \lt \int_1^x \tfrac1t \mathrm dt \lt \int_1^x t^{c1} \mathrm dt \\ 0 \lt \ln{x} \lt \tfrac1c x^c  1 \lt \tfrac1c x^c \\ \\ 0 \lt \ln^a{x} \lt \tfrac1c x^{ac} & (a \gt 0) \\ 0 \lt \frac{\ln^a{x}}{x^b} \lt \tfrac1c x^{acb} & (b \gt 0) \\ & \text{now, pick $c = \frac{b}{2a} \gt 0$} \\ 0 \lt \frac{\ln^a{x}}{x^b} \lt \tfrac1c x^{\frac{b}{2}} \\ \end{array}$$ 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 $b$. 

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