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burhan101
 2 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow 1} \sqrt{\frac{ 1t^2 }{ 1+t }}\]

kphilips2010
 2 years ago
Best ResponseYou've already chosen the best response.2why is your limit has x approahing 1 and the function is in terms of t?

burhan101
 2 years ago
Best ResponseYou've already chosen the best response.0that was a typo an my behalf :$

kphilips2010
 2 years ago
Best ResponseYou've already chosen the best response.2factor the numerator to (1+t)(1t) then cancel the (1+t) out so you are left with \[\sqrt{1t}\]

kphilips2010
 2 years ago
Best ResponseYou've already chosen the best response.2then you can substitue

burhan101
 2 years ago
Best ResponseYou've already chosen the best response.0@kphilips2010 that method is not correct

burhan101
 2 years ago
Best ResponseYou've already chosen the best response.0i dont think i ca use conjugates for this question

kphilips2010
 2 years ago
Best ResponseYou've already chosen the best response.2why is that wrong?

burhan101
 2 years ago
Best ResponseYou've already chosen the best response.0because it's not a binomia

kphilips2010
 2 years ago
Best ResponseYou've already chosen the best response.2they are not conjugates, they are addition and addition is commutative

kphilips2010
 2 years ago
Best ResponseYou've already chosen the best response.2I didnt say it was a binomial, I am saying you can factor the binomial into (1+t)(1t)

burhan101
 2 years ago
Best ResponseYou've already chosen the best response.0but isnt there another method, like multiply by the denominator or something of the like

kphilips2010
 2 years ago
Best ResponseYou've already chosen the best response.2no not in this case

kphilips2010
 2 years ago
Best ResponseYou've already chosen the best response.2LOL i am sure I am right...I hope

joemath314159
 2 years ago
Best ResponseYou've already chosen the best response.0kphilips method is good imo.

kphilips2010
 2 years ago
Best ResponseYou've already chosen the best response.2thanks joemath314159

burhan101
 2 years ago
Best ResponseYou've already chosen the best response.0i know thats how i approached the question at first but my teacher said this method was wrong ? :S

kphilips2010
 2 years ago
Best ResponseYou've already chosen the best response.2why did your teacher say you were wrong?

burhan101
 2 years ago
Best ResponseYou've already chosen the best response.0i dont know she said there was something wrong with the method :S

kphilips2010
 2 years ago
Best ResponseYou've already chosen the best response.2does she know what she is talking about? LOL

kphilips2010
 2 years ago
Best ResponseYou've already chosen the best response.2sorry but i dont have another solution

burhan101
 2 years ago
Best ResponseYou've already chosen the best response.0is this what you're saying dw:1350442853605:dw

burhan101
 2 years ago
Best ResponseYou've already chosen the best response.0so do i keep the square root sign ?

joemath314159
 2 years ago
Best ResponseYou've already chosen the best response.0If you want to be rigorous about it, you need to state that for limits, you are looking at values of t close to 1, but NOT EQUAL to 1. if t is not equal to 1, then:\[\frac{1t^2}{1+t}=\frac{(1t)(1+t)}{(1+t)}=1t\]We can only divide those monomials out if t DOESNT equal 1, which is ok in limits, we dont want t to equal 1. Im making an emphasis on the fact that if t was equal to 1, then that division doesnt make any sense.

burhan101
 2 years ago
Best ResponseYou've already chosen the best response.0ohh i see what you're saying

burhan101
 2 years ago
Best ResponseYou've already chosen the best response.0does the squareroot say there tho ?

kphilips2010
 2 years ago
Best ResponseYou've already chosen the best response.2yes the sq rt stay
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