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anonymous
 5 years ago
use l'hopital's rule to solve the limit as it approaches 0 from the right of 3/sinx  1/x ??
anonymous
 5 years ago
use l'hopital's rule to solve the limit as it approaches 0 from the right of 3/sinx  1/x ??

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[ \lim_{x \rightarrow 0+} 3/\sin x  1/x\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0limit is 0 l'hopitals rule you'll get lim of 0/cos(0)  0/1 so 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so 0 is also an indeternminate form?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks by the wayyyy!!! :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no, if the limit is 0, it exists it goes to 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if it comes out to be 0 / 0 then it's indeternminate

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0First show that limit is indeterminate, then employ l'hopital's rule lim x>0 (3xsinx)/(xsinx)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i'm not sure if cyter has it right... i'm pretty sure you deriv instead of integral for l'hopital's rule

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this function ends up equaling 0.. thats where i am confused bcuz why do i need to use l'hopitals rule/.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lim 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
 5 years ago
Best ResponseYou've already chosen the best response.0It's not 0. You can't use (and don't need) l'Hopital here.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0polpak, can you explain your reasoning?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Err wait. It's not 0, but you do need l'hopital

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how is my original equation (the question) undefined? 3 over sin(0)  1 over x = 3 over 0  1 over 0 = 0  0 = 0 ??

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0polpak when you get a chance will you come back to that kid who is trying to teach himself

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.03/0 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
 5 years ago
Best ResponseYou've already chosen the best response.0You don't have 0  0, you have \(3/0  1/0 = \infty  \infty\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0And you do need lhopital, but you need to first put over a common denominator

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ohhhh *faceplam* i forgot limits >< it's "as you approach" not zero thanks polpak yup itenn, polpak's right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[3/sinx  1/x = \frac{3x  sinx}{xsinx} \] Which gives a 0 over 0 and you take the derivative of top and bottom.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Give that derivative a try and take the limit as the derivative aproaches 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so far i have 3  cos x all over 2 cos x  x sinx

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I agree with the numerator, but not the denominator.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Using products rule \[\frac{d}{dx}x(sin\ x) = x(cos\ x) + (sin\ x)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Which gives \[\frac{3cosx}{x(cos\ x) + (sin\ x)}\] Which we can then plug in 0 for x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0why was the first equation indeterminate again.. im sorry im not getting this:(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it was similat to the 2/0+ ... it was 3/0  1/0.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Because you had 3/0  1/0. Which is \(\infty  \infty\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0And that can be a lot of different things. It could be 0, it could be a constant, or it could be +/ infinity.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So we have to put them over a common denominator to get 0/0 or \(\infty/\infty\) and then use l'Hopital's

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0to see which infinity is growing faster

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ohkay. i did another on my own  limit x approaches 1 [ x^3  1 all over x1 ]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i got 0/0 and then did derivative of top and bottom 3x^2 over 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and then i got 3. is that right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0my last problem for l'hopitals rule is limit x approaches 0+ x^x ??

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what is the indeterminate form... if 0^0 = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It'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
 5 years ago
Best ResponseYou've already chosen the best response.0So in this case you have \[\lim_{e^x \rightarrow 0} x(ln\ x)\] Which is of the form \(0\cdot\infty\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.00^0 is actually an indeterminate form so u can use l'hopital's rule

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and 0^0 is not equal to 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.00 * inf is an indeterminate form?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0polpk, I have a quick question on mine you were helping me with.. when you're done here:)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0http://en.wikipedia.org/wiki/Indeterminate_form#List_of_indeterminate_forms

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0inf/inf, 0/0, inf/inf, inf/inf or 0^0, any of those are indeterminate forms

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\(0 * \infty\) is indeterminate.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sometimes wikipedia is unreliable

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the next step then would be to take the derivativve?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but after futhre digging I'm convinced

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Was gonna say, wolfram alpha also thinks it's indeterminate: http://mathworld.wolfram.com/Indeterminate.html

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Which next step? I may have a different process for this than cyter.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im following you polpak..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'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
 5 years ago
Best ResponseYou've already chosen the best response.0thats where i confused.. l'hopitals rule has lim f(x)/g(x) and then f'(x)/g'(x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i dont see a fraction here.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Since \[\lim_{e^x \rightarrow 0} x(ln\ x)\] 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
 5 years ago
Best ResponseYou've already chosen the best response.0this one is very complicated.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i am not good with derivatices so the ln is trouble for me.. this is hard:(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So \[\frac{d}{du} ln\ 1/u = u*(\frac{1}{u^2}) = \frac{1}{u}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so now we have Erg, my notation is wrong, but it's tiny if I fix it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Anyway, 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
 5 years ago
Best ResponseYou've already chosen the best response.0So we have \[e^{\lim_{u \rightarrow \infty} \frac{1}{t}} = e^0 = 1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ohh it was never supposed to be e to the u

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, it was supposed to be e to the limit. That's it. The limit as \(x \rightarrow 0 \text{ of } x^x = 1\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so no l'hopital is needed??

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes it is. I used it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This is too difficult with such a tiny font. Lemme try something else

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0http://www.dabbleboard.com/draw?b=Guest665733&i=0&c=68e3fba29b21f18d6667a1c196154c7ed4b665ee
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