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evaluate the limit of x-8/ x^2-8 as x->8. I got 16.
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\[\frac{ 8-8 }{ 64-8 }=\frac{ 0 }{ 56 }\]
that's what I got I dont know how you got 16
\[\lim_{x\to 8} \frac{x-8}{x^2 - 8}\] Is this what you meant? then i must agree with recon14193 as x approaches 8 the denominator approaches 56, but the numerator approaches 0. The fraction tends to a finite value, which is zero.
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