You may come across limit problems in calculus that have cubic exponents in the numerator or denominator. The problem in taking limits when you’re given a cubic arises when the numerator and denominator of the function in question both equal zero if you plug in the number that the variable is being limited to. So you need to factor out and simplify the function to make the true limit more apparent.
Confirm that the function of concern in fact doesn’t have a clear limit if you just plug in the number being limited to.
For example, suppose you need to find the limit of [x^3-12x+16]/[x^2+2x-8] as x goes to 2. Here, the caret ^ indicates exponentiation. If you plug in 2, you get zero over zero, which doesn’t have any meaning.
Factor the numerator and denominator by dividing out a binomial that has as a root the number to which the variable converges in the limit.
This sounds complicated, so look at how that would apply to the example above. You’re taking the limit of the function above at 2. So divide the binomial x-2 into the cubic to get x^2+2x-8. (Note that [x^2+2x-8]*(x-2) = x^3-12x+16.) The function above is now (x-2)[x^2+2x-8]/[x^2+2x-8]. If you’ve forgotten polynomial long division from algebra class, see the Resources section below for an example.
Cancel out the common polynomials in the numerator and denominator.
Continuing with the example above, note that you didn’t really need to factor x-2 out of the denominator because the x^2+2x-8 in the numerator and the denominator cancel out, leaving x-2. But if you had factored x-2 out of the denominator, you’d have gotten (x-2)(x+4). So the function in its entirety would be (x-2)[x^2+2x-8]/[(x-2)(x+4)]. Canceling out the x-2 in the numerator and denominator leaves [x^2+2x-8]/(x+4).
Take the limit of the function now.
Continuing with the example, the limit as x goes to 2 of [x^2+2x-8]/(x+4) is 0 divided by 6 (i.e., 0).
Tips & Warnings
- To reduce the time that long division takes, two helpful equations for factoring cubics are these well-known formulas: (x-y)^3 = (x-y)(x^2+xy+y^2) and (x+y)^3 = (x+y)(x^2-xy+y^2). For example, (x-2)^3 can be written (x-2)(x^3+2x+4). How to remember such complex formulas? The negative sign in the binomial of the first equation is no surprise. From thereon, just keep in mind that the factorization for both equations has only one negative in it. So xy gets the negative sign in the other equation.
- Photo Credit BananaStock/BananaStock/Getty Images
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