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Amanderp Group Title

How does one handle the case of a limit that is in indeterminate form (turns out to be 0/0 when the number is plugged in)?

  • 2 years ago
  • 2 years ago

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  1. badreferences Group Title
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    The general solution amounts to applying l'Hopital's rule. Are you familiar?

    • 2 years ago
  2. Amanderp Group Title
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    No Im not. Can you show me?

    • 2 years ago
  3. badreferences Group Title
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    l'Hopital's rule asserts that, for any \(\lim_{x\to a}f(x)=\lim_{x\to a} g(x)=0,\infty\), the equivalence\[\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}\]can be established. For instance,\[\lim_{x\to0}\frac{\sin x}{x}=\lim_{x\to0}\frac{\frac{d}{dx}\sin(x)}{\frac{d}{dx}x}=\lim_{x\to0}\frac{\cos x}{1}=1\]

    • 2 years ago
  4. Amanderp Group Title
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    Oh. So essentially you're calculating the derivative of the initial problem and then finding the limit of that?

    • 2 years ago
  5. badreferences Group Title
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    You're finding the derivative of numerator and denominator (each evaluated separately), application-wise. The limits should be the same. This only applies when they normally evaluate to the indeterminate forms \(\frac00,\frac\infty\infty\).

    • 2 years ago
  6. Amanderp Group Title
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    So then this is sort of like the quotient rule in the sense that you have to break up the function into two separate problems and then evaluate to get the answer?

    • 2 years ago
  7. badreferences Group Title
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    Okay, I'll be less technical; sorry, been reading a textbook recently. Here's a practice problem for you.\[\lim_{x\to0}\frac{e^x-1}{x}=\,\,?\]

    • 2 years ago
  8. Amanderp Group Title
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    Okay. So. This is going to come out to 0/0 so you have to break it up right? So itll be the limit as it approaches 0 e^x-1 as one equation and the limit as it approaches 0 x, as the two problems?

    • 2 years ago
  9. badreferences Group Title
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    No, I'll break it down for you.\[\lim_{x\to0}\frac{e^x-1}{x}\]is our problem. We note that this evaluates to the indeterminate form \(\frac00\) by just plugging in \(0\) for \(x\). So, let's instead apply l'Hopital's rule. First, we need to make sure we meet the prerequisites: that \(\lim_{x\to0}e^x-1=\lim_{x\to0}x=0\text{ or }\infty\). Plugging in zero, as we just did, shows that this is the case. So, next step: apply the derivative of top and bottom.

    • 2 years ago
  10. Amanderp Group Title
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    Im sorry D: Im still a bit confused :/

    • 2 years ago
  11. Amanderp Group Title
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    Okay wait let me try continuing on paper. That explanation helped a bit.

    • 2 years ago
  12. badreferences Group Title
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    \[\lim_{x\to0}\frac{e^x-1}{x}\to\lim_{x\to0}\frac{\frac{d}{dx}(e^x-1)}{\frac{d}{dx}x}\]

    • 2 years ago
  13. badreferences Group Title
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    brb quick lunch. @Ishaan94 @eliassaab help here please

    • 2 years ago
  14. Ishaan94 Group Title
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    Amanderp? Is it still confusing you?

    • 2 years ago
  15. Amanderp Group Title
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    YES :(

    • 2 years ago
  16. Ishaan94 Group Title
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    What part is confusing you?

    • 2 years ago
  17. Amanderp Group Title
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    Im not sure. Like I know how to find the limit of something. But my teacher never actually went over this rule. And Im not really understanding it..

    • 2 years ago
  18. Amanderp Group Title
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    So I guess the rule itself, including the application is confusing me.

    • 2 years ago
  19. Ishaan94 Group Title
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    Hmm what you are supposed to do is simply differentiate the numerator and denominator, if the limit is of zero by zero or infinity by infinity form.

    • 2 years ago
  20. Amanderp Group Title
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    Im sorry. I need a refresh. Differentiation is simply taking the derivative right?

    • 2 years ago
  21. Ishaan94 Group Title
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    Yes.

    • 2 years ago
  22. Amanderp Group Title
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    Okay so then the answer to this problem that badreferences provided would be the derivative of the top and bottom, in fraction form?

    • 2 years ago
  23. Ishaan94 Group Title
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    Yes.

    • 2 years ago
  24. Amanderp Group Title
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    Like it wouldnt be evaluated, so to speak, with the 0. It would be left in terms of the variables?

    • 2 years ago
  25. Ishaan94 Group Title
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    It would be evaluated at the given limit. 1. Check the form. 2. If it's zero by zero or infinity by infinity, you apply L'Hospital. 3. Evaluate it for the given limit.

    • 2 years ago
  26. Amanderp Group Title
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    Okay but see thats what Im not sure about. Like I now kind of understand it, but I need a simpler definition of that rule. Because Im really not grasping it entirely.

    • 2 years ago
  27. Ishaan94 Group Title
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    I am sorry. I am not really a good teacher. I hope @eliassaab can help you.

    • 2 years ago
  28. Ishaan94 Group Title
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    I won't lie to you. Personally, I don't know why L'Hospital is used and what happens when you differentiate the numerator and denominator. For me it's more of a trick to get the limits quickly and I have always avoid using it.

    • 2 years ago
  29. Amanderp Group Title
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    Lol I appreciate your honesty. Im not a math person so its just not clicking :3

    • 2 years ago
  30. eliassaab Group Title
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    To avoid using L'Hospital's rule, you can sometimes factor up and down and cancel the term that is making up and down zero. \[ \lim_{x\to 7} \frac {x^2 - 8 x +7}{ x-7}=\frac 0 0\\ lim_{x\to 7} \frac {x^2 - 8 x +7}{ x-7}=lim_{x\to 7} \frac {(x-7)(x-1)}{ x-7}=\lim_{x\to 7}(x-1)=6 \]

    • 2 years ago
  31. Amanderp Group Title
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    Ohmygod thats ten times easier. Can you do that with all problems that would typically require the rule?

    • 2 years ago
  32. TuringTest Group Title
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    no, only in certain cases consider\[\lim_{x\to0}{\sin x\over x}\]this limit can be done with l'Hospital, but there is clearly no way to factor anything

    • 2 years ago
  33. Ishaan94 Group Title
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    \[\sin x \approx x\]For lower values of x. Maybe that's why \[\lim_{x\to0} \frac{\sin x}x = 1\]

    • 2 years ago
  34. badreferences Group Title
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    An approximation of \(\sin x\) at infinitesmal values around \(0\) reveals a slope of \(\cos0=1\therefore P(x)=x\).

    • 2 years ago
  35. eliassaab Group Title
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    Actually, the limit of sin(x)/x is used to prove that the derivative of sin(x) is cos(x). Technically, one should not use L'Hospital's rule to find the limit. See the proof on http://www.proofwiki.org/wiki/Limit_of_Sine_of_X_over_X/Geometric_Proof

    • 2 years ago
  36. eliassaab Group Title
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    See also this http://en.wikipedia.org/wiki/Differentiation_of_trigonometric_functions

    • 2 years ago
  37. badreferences Group Title
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    Interesting. Thanks for the correction. :)

    • 2 years ago
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