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\[\lim_{x \rightarrow -1} \sqrt{\frac{ 1-t^2 }{ 1+t }}\]
why is your limit has x approahing -1 and the function is in terms of t?
that was a typo an my behalf :$

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factor the numerator to (1+t)(1-t) then cancel the (1+t) out so you are left with \[\sqrt{1-t}\]
then you can substitue
@kphilips2010 that method is not correct
i dont think i ca use conjugates for this question
why is that wrong?
because it's not a binomia
binomial*
they are not conjugates, they are addition and addition is commutative
I didnt say it was a binomial, I am saying you can factor the binomial into (1+t)(1-t)
but isnt there another method, like multiply by the denominator or something of the like
no not in this case
LOL i am sure I am right...I hope
kphilips method is good imo.
thanks joemath314159
i know thats how i approached the question at first but my teacher said this method was wrong ? :S
hmmm I feel dumb
why did your teacher say you were wrong?
i dont know she said there was something wrong with the method :S
does she know what she is talking about? LOL
sorry but i dont have another solution
is this what you're saying |dw:1350442853605:dw|
so do i keep the square root sign ?
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:\[\frac{1-t^2}{1+t}=\frac{(1-t)(1+t)}{(1+t)}=1-t\]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.
ohh i see what you're saying
does the squareroot say there tho ?
yes.
yes the sq rt stay
yay i was right.
thanks :D

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