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

- anonymous

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

\[ \lim_{x \rightarrow 0+} 3/\sin x - 1/x\]

- anonymous

limit is 0
l'hopitals rule
you'll get
lim of 0/cos(0) - 0/1
so 0

- anonymous

so 0 is also an indeternminate form?

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- anonymous

thanks by the wayyyy!!! :)

- anonymous

no, if the limit is 0, it exists
it goes to 0

- anonymous

if it comes out to be 0 / 0
then it's indeternminate

- anonymous

First show that limit is indeterminate, then employ l'hopital's rule
lim x->0 (3x-sinx)/(xsinx)

- 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

this function ends up equaling 0.. thats where i am confused bcuz why do i need to use l'hopitals rule/.

- 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

It's not 0. You can't use (and don't need) l'Hopital here.

- anonymous

polpak, can you explain your reasoning?

- anonymous

Err wait. It's not 0, but you do need l'hopital

- 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

polpak when you get a chance will you come back to that kid who is trying to teach himself

- 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

You don't have 0 - 0, you have \(3/0 - 1/0 = \infty - \infty\)

- anonymous

And you do need lhopital, but you need to first put over a common denominator

- anonymous

ohhhh *faceplam*
i forgot
limits ><
it's "as you approach" not zero
thanks polpak
yup itenn, polpak's right

- 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

Give that derivative a try and take the limit as the derivative aproaches 0

- anonymous

I'll brb

- anonymous

thankk you!!!!

- anonymous

so far i have 3 - cos x all over 2 cos x - x sinx

- anonymous

I agree with the numerator, but not the denominator.

- anonymous

Using products rule
\[\frac{d}{dx}x(sin\ x) = x(cos\ x) + (sin\ x)\]

- anonymous

Which gives
\[\frac{3-cosx}{x(cos\ x) + (sin\ x)}\]
Which we can then plug in 0 for x

- anonymous

2/0 ?

- anonymous

Yep, \(2/0^+\)

- anonymous

So \(+\infty\)

- anonymous

why was the first equation indeterminate again.. im sorry im not getting this:(

- anonymous

it was similat to the 2/0+ ... it was 3/0 - 1/0.

- anonymous

Because you had 3/0 - 1/0. Which is \(\infty - \infty\)

- 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

So we have to put them over a common denominator to get 0/0 or \(\infty/\infty\) and then use l'Hopital's

- anonymous

to see which infinity is growing faster

- anonymous

ohkay.
i did another on my own -- limit x approaches 1 [ x^3 - 1 all over x-1 ]

- anonymous

i got 0/0 and then did derivative of top and bottom 3x^2 over 1

- anonymous

and then i got 3. is that right?

- anonymous

Yep, that's right.

- anonymous

my last problem for l'hopitals rule is limit x approaches 0+ x^x ??

- anonymous

what is the indeterminate form... if 0^0 = 0

- 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

So in this case you have
\[\lim_{e^x \rightarrow 0} x(ln\ x)\]
Which is of the form \(0\cdot\infty\)

- anonymous

0^0 is actually an indeterminate form so u can use l'hopital's rule

- anonymous

and 0^0 is not equal to 0

- anonymous

0 * inf is an indeterminate form?

- anonymous

no

- anonymous

polpk, I have a quick question on mine you were helping me with.. when you're done here:)

- anonymous

http://en.wikipedia.org/wiki/Indeterminate_form#List_of_indeterminate_forms

- anonymous

inf/inf, 0/0, -inf/inf, inf/-inf or 0^0, any of those are indeterminate forms

- anonymous

\(0 * \infty\) is indeterminate.

- anonymous

sometimes wikipedia is unreliable

- anonymous

the next step then would be to take the derivativve?

- anonymous

but after futhre digging
I'm convinced

- anonymous

Was gonna say, wolfram alpha also thinks it's indeterminate:
http://mathworld.wolfram.com/Indeterminate.html

- anonymous

Which next step? I may have a different process for this than cyter.

- anonymous

im following you polpak..

- 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

thats where i confused.. l'hopitals rule has lim f(x)/g(x) and then f'(x)/g'(x)

- anonymous

i dont see a fraction here.

- 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

this one is very complicated.

- anonymous

Yeah, a bit.

- anonymous

i am not good with derivatices so the ln is trouble for me.. this is hard:(

- anonymous

So \[\frac{d}{du} ln\ 1/u = u*(\frac{-1}{u^2}) = \frac{-1}{u}\]

- anonymous

ohhh okay.

- anonymous

so now we have
Erg, my notation is wrong, but it's tiny if I fix it.

- anonymous

what would change?

- 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

So we have
\[e^{\lim_{u \rightarrow \infty} \frac{-1}{t}} = e^0 = 1\]

- anonymous

See, tiny.

- anonymous

mm.. at which point?

- anonymous

ohh it was never supposed to be e to the u

- 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

so no l'hopital is needed??

- anonymous

yes it is. I used it.

- anonymous

This is too difficult with such a tiny font. Lemme try something else

- anonymous

ohkay.

- anonymous

i am confused:(

- anonymous

http://www.dabbleboard.com/draw?b=Guest665733&i=0&c=68e3fba29b21f18d6667a1c196154c7ed4b665ee

- anonymous

im there..

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