Question

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

Answer 1

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

Answer 2

limit is 0
l'hopitals rule
you'll get

lim of 0/cos(0) - 0/1
so 0

Answer 3

so 0 is also an indeternminate form?

Answer 4

thanks by the wayyyy!!! :)

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

i'm not sure if cyter has it right...
i'm pretty sure you deriv instead of integral for l'hopital's rule

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

polpak, can you explain your reasoning?

Answer 13

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

Answer 14

how is my original equation (the question) undefined?
3 over sin(0) - 1 over x = 3 over 0 - 1 over 0 = 0 - 0 = 0 ??

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

\[3/sinx - 1/x = \frac{3x - sinx}{xsinx} \]
Which gives a 0 over 0 and you take the derivative of top and bottom.

Answer 21

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

Answer 22

I'll brb

Answer 23

thankk you!!!!

Answer 24

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

Answer 25

I agree with the numerator, but not the denominator.

Answer 26

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

Answer 27

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

Answer 28

2/0 ?

Answer 29

Yep, \(2/0^+\)

Answer 30

So \(+\infty\)

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

And that can be a lot of different things. It could be 0, it could be a constant, or it could be +/- infinity.

Answer 35

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

Answer 36

to see which infinity is growing faster

Answer 37

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

Answer 38

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

Answer 39

and then i got 3. is that right?

Answer 40

Yep, that's right.

Answer 41

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

Answer 42

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

Answer 43

It's not 0. If you have \(\lim_{x \rightarrow c} f(x) = 0^0\) then you have to take the
\[\lim_{e^x \rightarrow c} ln\ f(x)\]

Answer 44

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

Answer 45

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

Answer 46

and 0^0 is not equal to 0

Answer 47

0 * inf is an indeterminate form?

Answer 48

no

Answer 49

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

Answer 50

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

Answer 51

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

Answer 52

\(0 * \infty\) is indeterminate.

Answer 53

sometimes wikipedia is unreliable

Answer 54

the next step then would be to take the derivativve?

Answer 55

but after futhre digging
I'm convinced

Answer 56

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

Answer 57

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

Answer 58

im following you polpak..

Answer 59

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

Answer 60

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

Answer 61

i dont see a fraction here.

Answer 62

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.

Answer 63

this one is very complicated.

Answer 64

Yeah, a bit.

Answer 65

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

Answer 66

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

Answer 67

ohhh okay.

Answer 68

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

Answer 69

what would change?

Answer 70

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.

Answer 71

So we have

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

Answer 72

See, tiny.

Answer 73

mm.. at which point?

Answer 74

ohh it was never supposed to be e to the u

Answer 75

Yeah, it was supposed to be e to the limit.
That's it.
The limit as \(x \rightarrow 0 \text{ of } x^x = 1\)

Answer 76

so no l'hopital is needed??

Answer 77

yes it is. I used it.

Answer 78

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

Answer 79

ohkay.

Answer 80

i am confused:(

Answer 81

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

Answer 82

im there..