y=lmx->inf (xe^x)/(x^2+2x-1)
use l hopitals rule

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- anonymous

y=lmx->inf (xe^x)/(x^2+2x-1)
use l hopitals rule

- schrodinger

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- zepdrix

Before using L'Hopital's Rule, verify that the limit is of an indeterminate form.
This limit appears to be of the indeterminate form:
\[\frac{ \infty }{ \infty }\]
So we can apply L'Hopital's Rule.
Do you need help applying the rule? :)

- anonymous

yeah i do pls

- anonymous

beacuse after applying the rule, i syill got inf/inf

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- zepdrix

|dw:1349037732010:dw|
Still getting Inf/Inf?
Good that is still an indeterminate form, which means you can apply L'Hop again! :)

- anonymous

so how do i apply it again?

- anonymous

after the 1st derivative, i got
e^x(1+x)/(2x+2)
then i plugged in inf and got inf/inf

- zepdrix

|dw:1349037925108:dw|
Yah you need to apply L'Hop from that spot again :)

- zepdrix

The important thing to remember is that -
You can only apply L'Hop rule a second time (or even third time, etc.. ) if it is STILL of an indeterminate form after the first application of L'Hop.
Still confused? :O

- anonymous

yeah i now understand thanks alot :)

- zepdrix

yay \c:/

- anonymous

hey so the answer will be?

- anonymous

infinity?

- zepdrix

So it looks like it simplifies to:\[\frac{ \infty }{ 2 }\]
Which, yes is simply Infinity. Good job! :)

- anonymous

hi i need a bit help with one other scenario as i wasnt in the class when the whole topic was thought
how about this
y=limx->0^+ x^2x

- anonymous

thats x approches 0 from the right

- zepdrix

|dw:1349039331256:dw|

- zepdrix

|dw:1349039504362:dw|

- zepdrix

|dw:1349039633140:dw|

- zepdrix

|dw:1349039718975:dw|

- zepdrix

|dw:1349039844810:dw|

- zepdrix

|dw:1349039949728:dw|
Get stuck on any part of that?
I tried to be detailed, since you missed the concept in class D':
It's a little tricky.

- anonymous

1?

- zepdrix

Yay \:D/

- anonymous

does this math change if the say x approaches 0 from the left??

- anonymous

0^-

- zepdrix

Yes.
The reason they made that distinction, is because the Natural log is only defined for values greater than 0.
If we were approaching 0 from the left, it wouldn't make any sense, because that value isn't in the domain of lnx.
So we are forced to approach from positive values! :)

- zepdrix

|dw:1349040248054:dw|

- anonymous

oh! ok thanks alot for your help.so this as far as l'hopitals rule goes, no other thing?
if is inderterminant u keep using the rule and taking the 2nd derivate

- zepdrix

Yes repeat as often as you are able to. :)
Important note: You need the CORRECT indeterminate form to apply L'Hop.
If I remember correctly you need one of these two forms:\[\frac{ 0 }{ 0 },\frac{ \infty }{ \infty }\]
In this problem we just did, we had the indeterminate form:
\[\infty*0\]So we had to move things around before we could apply L'Hop :)

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