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zepdrix Group TitleBest ResponseYou've already chosen the best response.1
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? :)
 2 years ago

krypton Group TitleBest ResponseYou've already chosen the best response.0
yeah i do pls
 2 years ago

krypton Group TitleBest ResponseYou've already chosen the best response.0
beacuse after applying the rule, i syill got inf/inf
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
dw:1349037732010:dw Still getting Inf/Inf? Good that is still an indeterminate form, which means you can apply L'Hop again! :)
 2 years ago

krypton Group TitleBest ResponseYou've already chosen the best response.0
so how do i apply it again?
 2 years ago

krypton Group TitleBest ResponseYou've already chosen the best response.0
after the 1st derivative, i got e^x(1+x)/(2x+2) then i plugged in inf and got inf/inf
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
dw:1349037925108:dw Yah you need to apply L'Hop from that spot again :)
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

krypton Group TitleBest ResponseYou've already chosen the best response.0
yeah i now understand thanks alot :)
 2 years ago

krypton Group TitleBest ResponseYou've already chosen the best response.0
hey so the answer will be?
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
So it looks like it simplifies to:\[\frac{ \infty }{ 2 }\] Which, yes is simply Infinity. Good job! :)
 2 years ago

krypton Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

krypton Group TitleBest ResponseYou've already chosen the best response.0
thats x approches 0 from the right
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
dw:1349039331256:dw
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
dw:1349039504362:dw
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
dw:1349039633140:dw
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
dw:1349039718975:dw
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
dw:1349039844810:dw
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
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.
 2 years ago

krypton Group TitleBest ResponseYou've already chosen the best response.0
does this math change if the say x approaches 0 from the left??
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
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! :)
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
dw:1349040248054:dw
 2 years ago

krypton Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
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 :)
 2 years ago
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