My Math Forum Need some help to solve this limit

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March 4th, 2019, 11:18 AM   #1
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Need some help to solve this limit

Hi everyone,

I have a problem that I don't know how to solve - ideally without using l'Hospital. Can you pls help?
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 March 5th, 2019, 06:38 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,685 Thanks: 2666 Math Focus: Mainly analysis and algebra First, writing $y=\frac1x$ we can see that there are actually two problems here: $$\lim_{x \to 0^-} \frac1x e^{-\frac1x} = \lim_{y \to -\infty} y e^{-y} \to (-\infty) \cdot \infty$$ So the function grows without bound in the negative direction. The second is $$\lim_{x \to 0^+} \frac1x e^{-\frac1x} = \lim_{y \to +\infty} y e^{-y} = \lim_{y \to +\infty} \frac{y}{ e^{y}} = 0$$ This is a standard result which you would normally be allowed to quote directly. Since there are two different values for these two limits, you should be able to determine the correct answer to the given problem. Alternatively, you could graph the function and come to the same conclusion. Thanks from topsquark Last edited by v8archie; March 5th, 2019 at 06:42 AM.
March 5th, 2019, 01:04 PM   #3
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Quote:
 Originally Posted by v8archie First, writing $y=\frac1x$ we can see that there are actually two problems here: $$\lim_{x \to 0^-} \frac1x e^{-\frac1x} = \lim_{y \to -\infty} y e^{-y} \to (-\infty) \cdot \infty$$ So the function grows without bound in the negative direction. The second is $$\lim_{x \to 0^+} \frac1x e^{-\frac1x} = \lim_{y \to +\infty} y e^{-y} = \lim_{y \to +\infty} \frac{y}{ e^{y}} = 0$$ This is a standard result which you would normally be allowed to quote directly. Since there are two different values for these two limits, you should be able to determine the correct answer to the given problem. Alternatively, you could graph the function and come to the same conclusion.
The original question is for positive x only.

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