anonymous
  • anonymous
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)?
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
The general solution amounts to applying l'Hopital's rule. Are you familiar?
anonymous
  • anonymous
No Im not. Can you show me?
anonymous
  • anonymous
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,\[\lim_{x\to0}\frac{\sin x}{x}=\lim_{x\to0}\frac{\frac{d}{dx}\sin(x)}{\frac{d}{dx}x}=\lim_{x\to0}\frac{\cos x}{1}=1\]

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anonymous
  • anonymous
Oh. So essentially you're calculating the derivative of the initial problem and then finding the limit of that?
anonymous
  • anonymous
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 \(\frac00,\frac\infty\infty\).
anonymous
  • anonymous
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?
anonymous
  • anonymous
Okay, I'll be less technical; sorry, been reading a textbook recently. Here's a practice problem for you.\[\lim_{x\to0}\frac{e^x-1}{x}=\,\,?\]
anonymous
  • anonymous
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?
anonymous
  • anonymous
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.
anonymous
  • anonymous
Im sorry D: Im still a bit confused :/
anonymous
  • anonymous
Okay wait let me try continuing on paper. That explanation helped a bit.
anonymous
  • anonymous
\[\lim_{x\to0}\frac{e^x-1}{x}\to\lim_{x\to0}\frac{\frac{d}{dx}(e^x-1)}{\frac{d}{dx}x}\]
anonymous
  • anonymous
brb quick lunch. @Ishaan94 @eliassaab help here please
anonymous
  • anonymous
Amanderp? Is it still confusing you?
anonymous
  • anonymous
YES :(
anonymous
  • anonymous
What part is confusing you?
anonymous
  • anonymous
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..
anonymous
  • anonymous
So I guess the rule itself, including the application is confusing me.
anonymous
  • anonymous
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.
anonymous
  • anonymous
Im sorry. I need a refresh. Differentiation is simply taking the derivative right?
anonymous
  • anonymous
Yes.
anonymous
  • anonymous
Okay so then the answer to this problem that badreferences provided would be the derivative of the top and bottom, in fraction form?
anonymous
  • anonymous
Yes.
anonymous
  • anonymous
Like it wouldnt be evaluated, so to speak, with the 0. It would be left in terms of the variables?
anonymous
  • anonymous
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.
anonymous
  • anonymous
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.
anonymous
  • anonymous
I am sorry. I am not really a good teacher. I hope @eliassaab can help you.
anonymous
  • anonymous
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.
anonymous
  • anonymous
Lol I appreciate your honesty. Im not a math person so its just not clicking :3
anonymous
  • anonymous
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. \[ \lim_{x\to 7} \frac {x^2 - 8 x +7}{ x-7}=\frac 0 0\\ lim_{x\to 7} \frac {x^2 - 8 x +7}{ x-7}=lim_{x\to 7} \frac {(x-7)(x-1)}{ x-7}=\lim_{x\to 7}(x-1)=6 \]
anonymous
  • anonymous
Ohmygod thats ten times easier. Can you do that with all problems that would typically require the rule?
TuringTest
  • TuringTest
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
anonymous
  • anonymous
\[\sin x \approx x\]For lower values of x. Maybe that's why \[\lim_{x\to0} \frac{\sin x}x = 1\]
anonymous
  • anonymous
An approximation of \(\sin x\) at infinitesmal values around \(0\) reveals a slope of \(\cos0=1\therefore P(x)=x\).
anonymous
  • anonymous
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
anonymous
  • anonymous
See also this http://en.wikipedia.org/wiki/Differentiation_of_trigonometric_functions
anonymous
  • anonymous
Interesting. Thanks for the correction. :)

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