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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
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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.
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Last edited by v8archie; March 5th, 2019 at 06:42 AM.
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March 5th, 2019, 01:04 PM   #3
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Quote:
Originally Posted by v8archie View Post
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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