anonymous
  • anonymous
Really need help understanding this limit/derivative question... The limit Lim x approachs 5pi CosX +1 / x-5pi represents the derivative of some function f(x) at some number a. Find f and a . I don't even understand the wording of this question...
Mathematics
katieb
  • katieb
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zepdrix
  • zepdrix
Remember the limit definition of a derivative?\[\large f'(x)=\lim_{h \rightarrow 0}\frac{f(x-h)-f(x)}{h}\]Well there is also another definition that we see less often, of this form,\[\large f'(x)=\lim_{x \rightarrow a}\frac{f(x)-f(a)}{x-a}\]This is the one that we want to analyze.
zepdrix
  • zepdrix
If we compare this to the limit we were given, we can see that is looks like our A value will be 5pi yes? \[\large \lim_{x \rightarrow 5\pi}\frac{\cos x +1}{x-5\pi}\]
zepdrix
  • zepdrix
The top is still a little tricky though, we have to sort it out.

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anonymous
  • anonymous
wait the second one just looks like the mean value theorem? it oddly looks like f(a)-b/a-b = f'(c)
anonymous
  • anonymous
oh nevermind me theres a limit sorry im a bit tired
zepdrix
  • zepdrix
heh, yah it looks similar :)
zepdrix
  • zepdrix
\[\large \cos(5\pi)=?\]
anonymous
  • anonymous
So yeah, but we're looking for the original function is that it or looking for the derivative? and cos5pi would be... .96 ? But I think i did it in degree so it might be wrong.
zepdrix
  • zepdrix
That's one of your special angles that you're going to want to remember. It will produce the same value as Pi. 5pi is Pi with an extra spin around the circle. -1 yes?
zepdrix
  • zepdrix
With 2 extra spins* my bad.
anonymous
  • anonymous
I'll ask you one thing before we continue, when taking pi in a derivative, do we ALWAYS use it in radians? I mean sometimes it works when I don't put it in my calculator as a radian
zepdrix
  • zepdrix
If you're dealing with Pi, then yes you need to be in radians :o You could convert to degrees if radians are confusing you though.
anonymous
  • anonymous
i mean pi i mean cos and sin... wow I' m sounding stupid
zepdrix
  • zepdrix
5pi is the same as 180 degrees.
anonymous
  • anonymous
so in the question am I using L'hospital rule or finding the limit? that's where I'm lost. the question is really confusing . And alright thanks for explaining the pi. :)
zepdrix
  • zepdrix
No L'Hop. We're relating a weird looking limit back to the Limit Definition of a Derivative. We need to match up the pieces so we can see what the original function was. So far we've established that our A value is 5pi. If you're unsure about that, compare the form of our limit with the Definition,\[\large \large f'(x)=\lim_{x \rightarrow a}\frac{f(x)-f(a)}{x-a}\qquad\qquad \rightarrow \qquad \qquad \lim_{x \rightarrow 5\pi}\frac{\cos x +1}{x-5\pi}\]
zepdrix
  • zepdrix
See the a?
anonymous
  • anonymous
Yeah it's represented by 5pi?
anonymous
  • anonymous
so i have to equate both limit? and finding my h?
zepdrix
  • zepdrix
no h, we're using the second definition that i posted, not the one involving h.
zepdrix
  • zepdrix
It's another form of the limit definition. It comes up less often.
anonymous
  • anonymous
My bad I understood we had to manipulate it back to the other form
anonymous
  • anonymous
Then when its ask to find F, am I suppose to find cosx?
anonymous
  • anonymous
sorry, the wording is really throwing me off
zepdrix
  • zepdrix
If we can show that the limit matches the DEFINITION, then we can show what our F is. So we've established that 5pi matches the A we're looking for. We've also shown that cos(5pi)=-1. If we can somehow find a -1 in the top of that fraction, we can make it look like the Definition.
anonymous
  • anonymous
Oh wow. I understand. This was a very basic question but I have never seen that definition in my entire class. Thanks a lot! :)
zepdrix
  • zepdrix
\[\large \frac{\cos x+1}{x-5\pi} \qquad =\qquad \frac{\cos x-(-1)}{x-5\pi} \qquad = \qquad \frac{\cos x-(\cos 5\pi)}{x-5\pi}\]
zepdrix
  • zepdrix
Make sense? :) k cool!
anonymous
  • anonymous
Thanks a lot you're patient!

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