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June 5th, 2017, 09:00 AM   #1
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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.
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June 5th, 2017, 11:11 AM   #2
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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
$$\lim_{t \to +\infty} \frac{t^a}{e^{bt}}=0 \quad \text{for all positive $a$ and $b$}$$
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June 5th, 2017, 11:17 AM   #3
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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.
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June 5th, 2017, 01:43 PM   #4
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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.
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June 5th, 2017, 02:24 PM   #5
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Yes.
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June 5th, 2017, 03:23 PM   #6
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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 $b$.
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