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factor the denominator to start
\[\lim_{x \rightarrow -3}{x+3\over x ^{2}-x-12}\]
\[x+1\over x ^{2}-x-4\] @ChukRock like this?
He meant to factor \(x^2-x-4\)
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no like this \[(x+3)/(x+3)(x-4)\]
that way you can cancel the (x+3) in both numerator and denominator
oh okay yea its gonna be \[-1\over 7\]
-4+3 = -1 -4*3 = -12 x^2 + 3x -4x -12 x(x+3)-4(x+3) (x-4)(x+3)
@ChukRock 's way is the more elegant, in my opinion. Another way of doing it is to divide everything by x^2.
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@bmp good seeing you again! oh okay thanks for the tip
thanks @zepp I appreciate the help
No problem, and same to you :-)
You are welcome :)
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