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lim x->16, x^2-256/(sqrt(x)-4)
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first factorize the top into (x+16)(x-16) (difference of squares) then factorize (x-16) into (sqrt(x)-4)(sqrt(x)+4) now cancel and substitute x=16 Answer is 256
\[{(x^2-256)(\sqrt{x}+4) \over (\sqrt{x}-4)(\sqrt{x}+4)}={(x-16)(x+16)(\sqrt{x}+4) \over x-16}=(x+16)(\sqrt{x}+4)\] Just plug x=16 and get your answer.
ya thats right ty i did not factor the top first
You're welcome!
how did you know to factor the top first?
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Because I knew that it would be cancelled out with the bottom.
haha ok lol i guess i should just do some more
Yeah I guess so :D
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