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?
thanks!
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.
To factor the numerator, you can start by seeing if (x-1) is a factor. You can substitute x = 1 into the expression and check if it equals zero. If it does, then (x-1) is a factor. If it doesn't, you can try (x+1) as a factor and repeat the process. Once you find a factor, you can use long division or synthetic division to divide the numerator by that factor and obtain the remaining factor.
Once you have factored the numerator and denominator, you can simplify the expression and then take the limit. In this case, you want to find the limit of the cube root of the expression as x approaches 3. So after factoring and simplifying, you can evaluate each limit separately, and then take the cube root of the quotient.
I hope this helps! Let me know if you have any further questions.