Limit Definition Derivative: e^(-1/x)

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Homework Statement



Let

f(x) =\begin{cases}
0 & \text{ if } x\leq 0 \\
e^\left ( -1/x \right ) & \text{ if } x> 0
\end{cases}
Compute f'(x) for x < 0 and x > 0.

Homework Equations



[tex] f'(x) = \lim \ \ \ \ \ \ \displaystyle{\frac{e^{(-1/(x+h)} - e^{-1/x}}{h}} \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ h\rightarrow0[/tex]

The Attempt at a Solution



Did the x < 0 part already, seemed straightforward (got 0/h = 0).

For x > 0:

[tex]f'(x) = \lim \ \ \ \ \ \ \frac{1}{he^{1/(x+h)}} - \ \ \ \ \lim \ \ \ \ \ \frac{1}{he^{1/x}} \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ h\rightarrow0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ h\rightarrow0[/tex]

It is it true that because, as h -> 0 the h in the exponential's denominator (in the first limit) goes to zero, the two limits are approximately the same (large) quantity, so their difference is zero?

EDIT: Oops, looks like I broke up the limit incorrectly, since it's not a limit of two separate functions of x. Should I instead take the limit of the difference of two quotients?
 
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Answers and Replies

  • #2
joshmccraney
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how have you defined e^x?? sometimes this definition is very helpful to solve
 
  • #3
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Do you mean I could use the Taylor series definition for e^x, one centered around 0 and one around h, to turn the exponentials into polynomials? Then the terms without h in the numerator would cancel, and I could use the remaining ones to cancel h in the denominator?
 
  • #4
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Or

[tex] e^{x} = \lim \ \ \ \ \ \ \ (1 + \frac{x}{n})^n \ \ \ ?\\ \ \ \ \ \ \ \ \ n\rightarrow∞[/tex]

EDIT: Then inserting -1/(x+h) for x, the exponential would go to 1, as would e^(-1/x)?
 
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  • #5
Dick
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Do you mean I could use the Taylor series definition for e^x, one centered around 0 and one around h, to turn the exponentials into polynomials? Then the terms without h in the numerator would cancel, and I could use the remaining ones to cancel h in the denominator?
Do you really have to do this using difference quotients? Why not just apply the chain rule to get the derivative for x>0? The only case where you might need to resort to the difference quotient is x=0.
 
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  • #6
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You may be right; I had gotten in the habit from using quotients in earlier parts of the question. I guess "compute" implies that using the chain rule is okay. Thanks for the suggestion. Out of curiosity, would one of the above methods be useful if I had to do it this way?
 
  • #7
joshmccraney
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sorry, judging by the time of the current academic term i was under the impression you needed to use the difference quotient. if you know the series, then why not use lopital's rule?
 
  • #8
Dick
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You may be right; I had gotten in the habit from using quotients in earlier parts of the question. I guess "compute" implies that using the chain rule is okay. Thanks for the suggestion. Out of curiosity, would one of the above methods be useful if I had to do it this way?
I suppose you can tough it out with the power series expansion. But I don't think you have to. Just use the chain rule for x>0. Your function is well behaved for x>0 and x<0. At x=0 it's different. It's the interesting point. 1/0 is undefined. There I think you need to apply the difference quotient. And as joshmccraney said, there you might find l'Hopitals rule helpful.
 
  • #9
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This problem has caught my attention.. I was thinking of using binomial expansion for e^(x+h)^-1 term and then to apply e^k -1 / k = 1 as lim k--> 0
 
  • #10
Dick
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This problem has caught my attention.. I was thinking of using binomial expansion for e^(x+h)^-1 term and then to apply e^k -1 / k = 1 as lim k--> 0
Sorry, but I'm not sure what part of the problem you trying to deal with here.
 
  • #11
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I was referring to the second part of the question. Suppose f(x) = e^(-1/x)

[tex] f'(x) = \lim_{h \to 0} e^{(1/x+h)} - e^{(1/x)} / h [/tex]

A taylor expansion about [tex] e^-{(1/x)} = \sum_{n=0} ^ \infty - (1/x^n) /n! [/tex]

[tex] f'(x) = \lim_{h \to 0} (-1 - 1/{(x+h)}) - (-1 - {1/x}) / h [/tex]

[tex] f'(x) = \lim_{h \to 0}{ {-(x+h)^{-1}} + (1/x) / h } [/tex]

Is this a reasonable approach ?
 
  • #12
Dick
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I was referring to the second part of the question. Suppose f(x) = e^(-1/x)

[tex] f'(x) = \lim_{h \to 0} e^{(1/x+h)} - e^{(1/x)} / h [/tex]

A taylor expansion about [tex] e^-{(1/x)} = \sum_{n=0} ^ \infty - (1/x^n) /n! [/tex]

[tex] f'(x) = \lim_{h \to 0} (-1 - 1/{(x+h)}) - (-1 - {1/x}) / h [/tex]

[tex] f'(x) = \lim_{h \to 0}{ {-(x+h)^{-1}} + (1/x) / h } [/tex]

Is this a reasonable approach ?
There are two problems here. The first is that it isn't right. There is no reason to ignore the higher power terms. And the second is that this whole thing is completely unnecessary. You know how to differentiate e^x and you know how to differentiate -1/x. Use difference quotients on those if you have to. Then use the chain rule. The only place you even need to think about using a difference quotient is at x=0 where the whole chain rule thing is undefined. How many times do I have to say this?
 
  • #13
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There are two problems here. The first is that it isn't right. There is no reason to ignore the higher power terms. And the second is that this whole thing is completely unnecessary. You know how to differentiate e^x and you know how to differentiate -1/x. Use difference quotients on those if you have to. Then use the chain rule. The only place you even need to think about using a difference quotient is at x=0 where the whole chain rule thing is undefined. How many times do I have to say this?
Thanks for the explanation, apologies if I have annoyed you.
 
  • #14
Dick
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Thanks for the explanation, apologies if I have annoyed you.
That's ok, but the only real issue in this problem that takes thinking about is whether the function is differentiable at x=0. And everybody seems to be missing that. That's what's annoying.
 
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  • #15
D H
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That's ok, but the only real issue in this problem that takes thinking about is whether the function is differentiable at x=0. And everybody seems to be missing that. That's what's annoying.
Why is that so annoying? This is an introductory calculus problem, and the problem statement specifically avoids the issue of behavior t x=0. It asks the student to "Compute f'(x) for x < 0 and x > 0."


That said, this function is differentiable at x=0 in the sense that ##\lim_{x\to 0^-} f'(x) = \lim_{x\to 0^+} f'(x) = 0##.
 
  • #16
Dick
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Why is that so annoying? This is an introductory calculus problem, and the problem statement specifically avoids the issue of behavior t x=0. It asks the student to "Compute f'(x) for x < 0 and x > 0."


That said, this function is differentiable at x=0 in the sense that ##\lim_{x\to 0^-} f'(x) = \lim_{x\to 0^+} f'(x) = 0##.
Ok, so now I'm annoyed with myself for seeing a different problem than was intended. Sorry! And the function isn't just differentiable in the sense of that limit. The difference quotient also shows f'(0)=0. I'm guessing that might be in another part of the same question.
 

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