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Amanderp

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)?

  • one year ago
  • one year ago

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

    • one year ago
  2. Amanderp
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    No Im not. Can you show me?

    • one year ago
  3. badreferences
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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,\[\lim_{x\to0}\frac{\sin x}{x}=\lim_{x\to0}\frac{\frac{d}{dx}\sin(x)}{\frac{d}{dx}x}=\lim_{x\to0}\frac{\cos x}{1}=1\]

    • one year ago
  4. Amanderp
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    Oh. So essentially you're calculating the derivative of the initial problem and then finding the limit of that?

    • one year ago
  5. badreferences
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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 \(\frac00,\frac\infty\infty\).

    • one year ago
  6. Amanderp
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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?

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

    • one year ago
  8. Amanderp
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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?

    • one year ago
  9. badreferences
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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.

    • one year ago
  10. Amanderp
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    Im sorry D: Im still a bit confused :/

    • one year ago
  11. Amanderp
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    Okay wait let me try continuing on paper. That explanation helped a bit.

    • one year ago
  12. badreferences
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    \[\lim_{x\to0}\frac{e^x-1}{x}\to\lim_{x\to0}\frac{\frac{d}{dx}(e^x-1)}{\frac{d}{dx}x}\]

    • one year ago
  13. badreferences
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    brb quick lunch. @Ishaan94 @eliassaab help here please

    • one year ago
  14. Ishaan94
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    Amanderp? Is it still confusing you?

    • one year ago
  15. Amanderp
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    YES :(

    • one year ago
  16. Ishaan94
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    What part is confusing you?

    • one year ago
  17. Amanderp
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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..

    • one year ago
  18. Amanderp
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    So I guess the rule itself, including the application is confusing me.

    • one year ago
  19. Ishaan94
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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.

    • one year ago
  20. Amanderp
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    Im sorry. I need a refresh. Differentiation is simply taking the derivative right?

    • one year ago
  21. Ishaan94
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    Yes.

    • one year ago
  22. Amanderp
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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?

    • one year ago
  23. Ishaan94
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    Yes.

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

    • one year ago
  25. Ishaan94
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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.

    • one year ago
  26. Amanderp
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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.

    • one year ago
  27. Ishaan94
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    I am sorry. I am not really a good teacher. I hope @eliassaab can help you.

    • one year ago
  28. Ishaan94
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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.

    • one year ago
  29. Amanderp
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    Lol I appreciate your honesty. Im not a math person so its just not clicking :3

    • one year ago
  30. eliassaab
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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. \[ \lim_{x\to 7} \frac {x^2 - 8 x +7}{ x-7}=\frac 0 0\\ lim_{x\to 7} \frac {x^2 - 8 x +7}{ x-7}=lim_{x\to 7} \frac {(x-7)(x-1)}{ x-7}=\lim_{x\to 7}(x-1)=6 \]

    • one year ago
  31. Amanderp
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    Ohmygod thats ten times easier. Can you do that with all problems that would typically require the rule?

    • one year ago
  32. TuringTest
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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

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

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

    • one year ago
  35. eliassaab
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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

    • one year ago
  36. eliassaab
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    See also this http://en.wikipedia.org/wiki/Differentiation_of_trigonometric_functions

    • one year ago
  37. badreferences
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    Interesting. Thanks for the correction. :)

    • one year ago
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