SOLUTION: Algebraically solve for the limit as x approaches 0 of ((2+x)^3 - 8)/x

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Question 218402: Algebraically solve for the limit as x approaches 0 of ((2+x)^3 - 8)/x
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
With "easy" limits you just replace the variable with the number it approaches. But we can't do that here because we end up with a zero in the denominator.

So what we have to do is try to simplify the fraction and hope that we end up a denominator which will not approach zero.

Simplifying/reducing fractions involves canceling common factors. So we need to factor the numerator and denominator. We could multiply out the the first part of the numerator, add like terms and then try to factor what remains.

But it is much easier if we realize that we can factor the numerator as it is now! The numerator is a difference of cubes, %282%2Bx%29%5E3+-+8+=+%282%2Bx%29%5E3+-+%282%29%5E3, and so it will factor according to the difference of cubes pattern: a%5E3+-+b%5E3+=+%28a-b%29%28a%5E2%2Bab%2Bb%5E2%29. Factoring this way we get:

We can see that the x's will cancel. (Note: we are finding the limit as x approaches 0. This means that x is never actually zero. So we are not canceling 0/0 (which is undefined). We are canceling (very small fraction)/(very small fraction) which is a 1.). This leaves us with:
%282%2Bx%29%5E2+%2B+%282%2Bx%292+%2B+4
and the limit of this as x approaches zero is one of the "easy" limits. Just replace the x's with 0's:
%282%2B0%29%5E2+%2B+%282%2B0%292+%2B+4+=+2%5E2+%2B+2%282%29+%2B+4+=+4+%2B+4+%2B+4+=+12