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krypton
y=lmx->inf (xe^x)/(x^2+2x-1) use l hopitals rule
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? :)
beacuse after applying the rule, i syill got inf/inf
|dw:1349037732010:dw| Still getting Inf/Inf? Good that is still an indeterminate form, which means you can apply L'Hop again! :)
so how do i apply it again?
after the 1st derivative, i got e^x(1+x)/(2x+2) then i plugged in inf and got inf/inf
|dw:1349037925108:dw| Yah you need to apply L'Hop from that spot again :)
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
yeah i now understand thanks alot :)
hey so the answer will be?
So it looks like it simplifies to:\[\frac{ \infty }{ 2 }\] Which, yes is simply Infinity. Good job! :)
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
thats x approches 0 from the right
|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.
does this math change if the say x approaches 0 from the left??
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! :)
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
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 :)