posted by . .
the limit of cuberoot((-3x^3+5x+2)/(x^2-1)) as x approaches 3 is the problem.
how could (-3x^3+5x+2) so it would be factored out with denominator?
If you read the answer I gave for the previous question, then you can take the limit inside to get
lim x->3 of((-3x^3+5x+2)/(x^2-1))^(1/3) becomes
(lim x->3(-3x^3+5x+2)/lim x->3(x^2-1))^(1/3) becomes
(-3(lim x->3)^3+5(lim x->3)+2)/((lim x->3)^2-1))^(1/3) =
(-81+15+2)/(9-1) = something you can do.
Be sure to check my arithmetic.
As for factoring it, you should see that the only factors of the denominator are (x-1) and (x+1). If neither of these divide the numerator, then no further simplifying can be done. (I think one of them should divide it by inspection, but I'll leave the work to you. Write back if you need help or want work checked.)
this one has to be cubed too, is that right?
No, I think I read this one correctly. You have, "the limit of cuberoot[epression]..."
So we will take the limit inside the function, evaluate it, then take the cube-root. The only difference here is that we have the 3rd root of a rational function -a function that is the ratio of two polynomials. Other than that, we can still take the limit inside of each function and evaluate them separately, then take roots.