anonymous
  • anonymous
use l'hopital's rule to solve the limit as it approaches 0 from the right of 3/sinx - 1/x ??
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[ \lim_{x \rightarrow 0+} 3/\sin x - 1/x\]
anonymous
  • anonymous
limit is 0 l'hopitals rule you'll get lim of 0/cos(0) - 0/1 so 0
anonymous
  • anonymous
so 0 is also an indeternminate form?

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anonymous
  • anonymous
thanks by the wayyyy!!! :)
anonymous
  • anonymous
no, if the limit is 0, it exists it goes to 0
anonymous
  • anonymous
if it comes out to be 0 / 0 then it's indeternminate
anonymous
  • anonymous
First show that limit is indeterminate, then employ l'hopital's rule lim x->0 (3x-sinx)/(xsinx)
anonymous
  • anonymous
i'm not sure if cyter has it right... i'm pretty sure you deriv instead of integral for l'hopital's rule
anonymous
  • anonymous
this function ends up equaling 0.. thats where i am confused bcuz why do i need to use l'hopitals rule/.
anonymous
  • anonymous
lim of 0/cos(0) - 0/1 ends up zero so the limit is zero but your original equation (the question) is undefined (or indeter.) so you have to apply l'hoptials' rule
anonymous
  • anonymous
It's not 0. You can't use (and don't need) l'Hopital here.
anonymous
  • anonymous
polpak, can you explain your reasoning?
anonymous
  • anonymous
Err wait. It's not 0, but you do need l'hopital
anonymous
  • anonymous
how is my original equation (the question) undefined? 3 over sin(0) - 1 over x = 3 over 0 - 1 over 0 = 0 - 0 = 0 ??
myininaya
  • myininaya
polpak when you get a chance will you come back to that kid who is trying to teach himself
anonymous
  • anonymous
3/0 is not 0 it's undefined. which is why i would do l'hopitals rule but polpak says i'm wrong :x so we'll see what he says first
anonymous
  • anonymous
You don't have 0 - 0, you have \(3/0 - 1/0 = \infty - \infty\)
anonymous
  • anonymous
And you do need lhopital, but you need to first put over a common denominator
anonymous
  • anonymous
ohhhh *faceplam* i forgot limits >< it's "as you approach" not zero thanks polpak yup itenn, polpak's right
anonymous
  • anonymous
\[3/sinx - 1/x = \frac{3x - sinx}{xsinx} \] Which gives a 0 over 0 and you take the derivative of top and bottom.
anonymous
  • anonymous
Give that derivative a try and take the limit as the derivative aproaches 0
anonymous
  • anonymous
I'll brb
anonymous
  • anonymous
thankk you!!!!
anonymous
  • anonymous
so far i have 3 - cos x all over 2 cos x - x sinx
anonymous
  • anonymous
I agree with the numerator, but not the denominator.
anonymous
  • anonymous
Using products rule \[\frac{d}{dx}x(sin\ x) = x(cos\ x) + (sin\ x)\]
anonymous
  • anonymous
Which gives \[\frac{3-cosx}{x(cos\ x) + (sin\ x)}\] Which we can then plug in 0 for x
anonymous
  • anonymous
2/0 ?
anonymous
  • anonymous
Yep, \(2/0^+\)
anonymous
  • anonymous
So \(+\infty\)
anonymous
  • anonymous
why was the first equation indeterminate again.. im sorry im not getting this:(
anonymous
  • anonymous
it was similat to the 2/0+ ... it was 3/0 - 1/0.
anonymous
  • anonymous
Because you had 3/0 - 1/0. Which is \(\infty - \infty\)
anonymous
  • anonymous
And that can be a lot of different things. It could be 0, it could be a constant, or it could be +/- infinity.
anonymous
  • anonymous
So we have to put them over a common denominator to get 0/0 or \(\infty/\infty\) and then use l'Hopital's
anonymous
  • anonymous
to see which infinity is growing faster
anonymous
  • anonymous
ohkay. i did another on my own -- limit x approaches 1 [ x^3 - 1 all over x-1 ]
anonymous
  • anonymous
i got 0/0 and then did derivative of top and bottom 3x^2 over 1
anonymous
  • anonymous
and then i got 3. is that right?
anonymous
  • anonymous
Yep, that's right.
anonymous
  • anonymous
my last problem for l'hopitals rule is limit x approaches 0+ x^x ??
anonymous
  • anonymous
what is the indeterminate form... if 0^0 = 0
anonymous
  • anonymous
It's not 0. If you have \(\lim_{x \rightarrow c} f(x) = 0^0\) then you have to take the \[\lim_{e^x \rightarrow c} ln\ f(x)\]
anonymous
  • anonymous
So in this case you have \[\lim_{e^x \rightarrow 0} x(ln\ x)\] Which is of the form \(0\cdot\infty\)
anonymous
  • anonymous
0^0 is actually an indeterminate form so u can use l'hopital's rule
anonymous
  • anonymous
and 0^0 is not equal to 0
anonymous
  • anonymous
0 * inf is an indeterminate form?
anonymous
  • anonymous
no
anonymous
  • anonymous
polpk, I have a quick question on mine you were helping me with.. when you're done here:)
anonymous
  • anonymous
http://en.wikipedia.org/wiki/Indeterminate_form#List_of_indeterminate_forms
anonymous
  • anonymous
inf/inf, 0/0, -inf/inf, inf/-inf or 0^0, any of those are indeterminate forms
anonymous
  • anonymous
\(0 * \infty\) is indeterminate.
anonymous
  • anonymous
sometimes wikipedia is unreliable
anonymous
  • anonymous
the next step then would be to take the derivativve?
anonymous
  • anonymous
but after futhre digging I'm convinced
anonymous
  • anonymous
Was gonna say, wolfram alpha also thinks it's indeterminate: http://mathworld.wolfram.com/Indeterminate.html
anonymous
  • anonymous
Which next step? I may have a different process for this than cyter.
anonymous
  • anonymous
im following you polpak..
anonymous
  • anonymous
I'm not sure how he was gonna use l'hopital on 0^0 since there's no fraction there, but I'm a bit rusty with this
anonymous
  • anonymous
thats where i confused.. l'hopitals rule has lim f(x)/g(x) and then f'(x)/g'(x)
anonymous
  • anonymous
i dont see a fraction here.
anonymous
  • anonymous
Since \[\lim_{e^x \rightarrow 0} x(ln\ x)\] is not nice, we have to swap our variable around a bit and say u = 1/x if x goes to 0, u goes to infinity. And that means that x = 1/u, so we have now \[\lim_{e^{u} \rightarrow \infty} \frac{ln\ 1/u}{u}\] Which gives us a nice fraction with \(\infty/\infty\) that we can take the derivative of.
anonymous
  • anonymous
this one is very complicated.
anonymous
  • anonymous
Yeah, a bit.
anonymous
  • anonymous
i am not good with derivatices so the ln is trouble for me.. this is hard:(
anonymous
  • anonymous
So \[\frac{d}{du} ln\ 1/u = u*(\frac{-1}{u^2}) = \frac{-1}{u}\]
anonymous
  • anonymous
ohhh okay.
anonymous
  • anonymous
so now we have Erg, my notation is wrong, but it's tiny if I fix it.
anonymous
  • anonymous
what would change?
anonymous
  • anonymous
Anyway, we have It's supposed to be e raised to the power of the limit of that whole thing, not e^u approaches infinity.
anonymous
  • anonymous
So we have \[e^{\lim_{u \rightarrow \infty} \frac{-1}{t}} = e^0 = 1\]
anonymous
  • anonymous
See, tiny.
anonymous
  • anonymous
mm.. at which point?
anonymous
  • anonymous
ohh it was never supposed to be e to the u
anonymous
  • anonymous
Yeah, it was supposed to be e to the limit. That's it. The limit as \(x \rightarrow 0 \text{ of } x^x = 1\)
anonymous
  • anonymous
so no l'hopital is needed??
anonymous
  • anonymous
yes it is. I used it.
anonymous
  • anonymous
This is too difficult with such a tiny font. Lemme try something else
anonymous
  • anonymous
ohkay.
anonymous
  • anonymous
i am confused:(
anonymous
  • anonymous
http://www.dabbleboard.com/draw?b=Guest665733&i=0&c=68e3fba29b21f18d6667a1c196154c7ed4b665ee
anonymous
  • anonymous
im there..

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