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

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

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

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

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

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

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

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

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

burhan101 Group TitleBest ResponseYou've already chosen the best response.0
binomial*
 2 years ago

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

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

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

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

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

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

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

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

kphilips2010 Group TitleBest ResponseYou've already chosen the best response.2
hmmm I feel dumb
 2 years ago

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

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

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

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

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

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

joemath314159 Group TitleBest ResponseYou've already chosen the best response.0
If 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.
 2 years ago

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

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

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

kphilips2010 Group TitleBest ResponseYou've already chosen the best response.2
yay i was right.
 2 years ago

burhan101 Group TitleBest ResponseYou've already chosen the best response.0
thanks :D
 2 years ago
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