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

burhan101 Group TitleBest ResponseYou've already chosen the best response.0
that was a typo an my behalf :$
 one year 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 \[\sqrt{1t}\]
 one year ago

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

joemath314159 Group TitleBest ResponseYou've already chosen the best response.0
yes.
 one year ago

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

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

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