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amyna
 one year ago
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?
amyna
 one year ago
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?

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phi
 one year ago
Best ResponseYou've already chosen the best response.3you could find the limit of the ln of that expression then the limit of the original will be e^(limit)

phi
 one year ago
Best ResponseYou've already chosen the best response.3yes and write that as \[ 2\frac{\ln x}{x^{1}} \]

amyna
 one year ago
Best ResponseYou've already chosen the best response.0ok thats all i needed to know! Thank You!

dan815
 one year ago
Best ResponseYou've already chosen the best response.0probably a better idea to do 2x/(1/lnx)

amyna
 one year ago
Best ResponseYou've already chosen the best response.0ya that seems much easier

dan815
 one year ago
Best ResponseYou've already chosen the best response.0the other way simplfies better :)

amyna
 one year ago
Best ResponseYou've already chosen the best response.0i got 0 so do i do the l hospitals rule again?

dan815
 one year ago
Best ResponseYou've already chosen the best response.0ya thats right now apply e^0= 1
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