amyna
  • amyna
ok so for this limit it is an infinity over an infinity. But how do i use l hospitals rule for this since its not in fraction form?
Mathematics
katieb
  • katieb
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amyna
  • amyna
|dw:1444587028789:dw|
phi
  • phi
you could find the limit of the ln of that expression then the limit of the original will be e^(limit)
amyna
  • amyna
so i could do 2x ln x ?

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phi
  • phi
yes and write that as \[ 2\frac{\ln x}{x^{-1}} \]
amyna
  • amyna
ok thats all i needed to know! Thank You!
dan815
  • dan815
probably a better idea to do 2x/(1/lnx)
amyna
  • amyna
ya that seems much easier
dan815
  • dan815
the other way simplfies better :)
amyna
  • amyna
i got 0 so do i do the l hospitals rule again?
dan815
  • dan815
ya thats right now apply e^0= 1
amyna
  • amyna
alrighty!!! Thanks!

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