Limits by algebra: factoring and rationalizing

When you reach for it

The limit laws turn most limits into arithmetic: for a polynomial p, \lim_{x \to a} p(x) = p(a), and for a quotient of polynomials whose denominator is not zero at a, the limit is the value of the quotient at a. They say nothing when the denominator heads to 0, and that case is not rare. Every derivative in calculus is a limit of a quotient whose top and bottom both shrink to 0, so it is the case that matters most.

Try the limit

Substituting x = 4 gives \displaystyle \frac{16 - 4 - 12}{4 - 4} = \frac{0}{0}. That is not an answer and not a statement that the limit fails. It is a report that the quotient laws do not apply, and the formula must be rewritten before they can. This lesson rewrites it in two ways: by factoring and canceling a common factor (part 1), and by rationalizing with a conjugate when a square root is involved (part 2).

The lesson takes two days. Part 1 is the reading of a substitution, why canceling is allowed, and factoring; Examples 1 to 4 go with it. Part 2 is rationalizing; Examples 5 to 7 go with it.

The procedure

Substitute first, and read what comes out

Substitute x = a into the numerator and the denominator separately. Three things can happen.

The middle row deserves a reason. If the top approaches a number c \ne 0 while the bottom approaches 0, the quotient is c divided by tiny numbers, so its size grows without bound. It cannot settle at any height, so the limit does not exist as a number. Canceling cannot rescue it: a common factor x - a would have to make the numerator 0 at a, and it is not. How such a quotient blows up, and on which side, is the subject of Infinite limits and vertical asymptotes.

The last row is called an indeterminate form, because \displaystyle \frac{0}{0} alone does not determine the limit. The quotients \displaystyle \frac{3x}{x}, \displaystyle \frac{x^3}{x} and \displaystyle \frac{x^2}{x^4} are all of this form as x \to 0: the first has limit 3, the second has limit 0, and the third, equal to \displaystyle \frac{1}{x^2} for x \ne 0, grows without bound and has no limit.

Why canceling is allowed

For x \ne 4,

because dividing by x - 4 is allowed whenever x - 4 \ne 0. At x = 4 the two sides are not equal: the left side is undefined and the right side is 7. So canceling does not produce the same function. It produces a function that agrees with it everywhere except at 4.

That is enough, by one principle.

Agreement principle. If f(x) = g(x) for every x near a other than a itself, and \lim_{x \to a} g(x) exists, then \lim_{x \to a} f(x) exists and equals it.

The reason is in the meaning of a limit from What a limit is: graphs and tables: \lim_{x \to a} f(x) = L says the outputs f(x) come as close to L as you like for all inputs close enough to a, with x \ne a. The value at a, or its absence, never enters. Two functions that match at every input the limit looks at have the same limit. (The careful version of “close enough” is the epsilon delta definition; the principle follows from it in one line.)

So

where the last step is direct substitution into a polynomial. The graph shows it: the line y = x + 3 with one point removed.

Why zero over zero signals a common factor

If a polynomial p has p(a) = 0, then by the The remainder and factor theorems x - a is a factor of p. So when a quotient of polynomials gives \displaystyle \frac{0}{0} at a, both top and bottom contain the factor x - a, and it can always be found and canceled. Finding it uses the factoring you already know: factoring trinomials, the difference of squares, and the cube patterns

After canceling, substitute again. If the form is still \displaystyle \frac{0}{0}, the factor appeared more than once; cancel again.

Rationalizing when a root is involved

A difference such as \displaystyle \sqrt{x + 4} - 3 is not a polynomial, so the factor theorem does not apply to it directly. The repair is the conjugate from Rationalizing a denominator: multiply by \displaystyle \sqrt{x + 4} + 3, and the product of the conjugates is a difference of squares with no root left,

In that lesson the conjugate cleared a root from a denominator; here it usually clears one from the numerator, and the point is not neatness but the factor x - a that appears. Multiplying top and bottom by the conjugate multiplies by 1 wherever the conjugate is not 0. Since \displaystyle \sqrt{x + 4} + 3 \ge 3, that is everywhere in the domain, so the rewritten quotient equals the original wherever the original is defined.

Worked examples

Boundaries

Canceling is for the form \displaystyle \frac{0}{0} only. If the denominator is not 0 at a, substitute; the answer is already there. If the numerator is not 0 and the denominator is, there is no common factor x - a, and the limit does not exist as a number. The one-sided behavior there belongs to Infinite limits and vertical asymptotes.

The simplified formula is not the original function. \displaystyle \frac{(x - 4)(x + 3)}{x - 4} and x + 3 differ at x = 4. The limit of the first is 7; its value at 4 is still undefined. Writing f(4) = 7 is false. When a limit exists but the value does not, the function has a removable break, taken up in Continuity and Discontinuities and how to remove them.

The last substitution needs a function that substitution works for. After canceling, direct substitution is justified by the limit laws only because the simplified expression is a polynomial, a quotient with nonzero denominator at a, or a root of something positive at a. If it is none of these, the work is not finished.

The conjugate needs the roots defined near a. If a root is defined on only one side of a, as \displaystyle \sqrt{x - 5} is near 5, only a one-sided limit can be asked; see One sided limits and when a limit exists. A conjugate pair also works only for square roots. A cube root needs the cube pattern a^3 - b^3 = (a - b)(a^2 + ab + b^2) instead, as in Exercise 25.

Not every \displaystyle \frac{0}{0} yields to algebra. \displaystyle \frac{\sin x}{x} has no factor to cancel; its limit at 0 needs a geometric argument in The squeeze theorem and the limit of sine x over x. Later, L’Hopital’s rule handles such quotients with derivatives.

Where this goes next

The difference quotients of Example 4 are the start of Finding a derivative from the definition: the derivative of f at a is \displaystyle \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}, which always begins as \displaystyle \frac{0}{0}, and for polynomials, reciprocals and square roots it is evaluated with exactly the expanding, combining and rationalizing practiced here. The derivative of \displaystyle \sqrt{x}, for instance, is a conjugate computation.

A limit that exists where the function does not is the mark of a removable discontinuity; defining the missing value as the limit repairs the function, which is how Continuity uses this lesson. And when algebra runs out, as for \displaystyle \frac{\sin x}{x}, the squeeze theorem and later L’Hopital’s rule take over the same question.

Exercises

Day 1: factoring and canceling

Evaluate each limit.

1. \lim_{x \to -3} \dfrac{x^2 + 7x + 12}{x + 3}
2. \lim_{x \to 6} \dfrac{x^2 - 36}{x^2 - 4x - 12}
3. \displaystyle \lim_{x \to \frac{1}{2}} \dfrac{2x^2 + 3x - 2}{2x - 1}
4. \lim_{x \to -1} \dfrac{x^3 + 1}{x + 1}
5. \lim_{x \to 1} \dfrac{x^4 - 1}{x^3 - 1}
6. \lim_{x \to 2} \dfrac{x^2 - 4x + 4}{x^2 - x - 2}

Day 1: read the substitution first

Let k(x) = \dfrac{x^2 - 5x + 4}{x^2 - x - 12}. Find each limit or explain why it does not exist.

7. \lim_{x \to 4} k(x)
8. \lim_{x \to 1} k(x)
9. \lim_{x \to -3} k(x)
10. \lim_{x \to 0} k(x)

Day 1: expand or combine first

11. \lim_{h \to 0} \dfrac{(h - 1)^3 + 1}{h}
12. \lim_{h \to 0} \dfrac{(2 + h)^2 + 5(2 + h) - 14}{h}
13. \displaystyle \lim_{x \to 2} \dfrac{\frac{1}{x + 1} - \frac{1}{3}}{x - 2}

Day 2: rationalizing

14. \displaystyle \lim_{x \to 0} \dfrac{\sqrt{1 + x} - 1}{x}
15. \displaystyle \lim_{x \to 3} \dfrac{\sqrt{x + 1} - 2}{x - 3}
16. \displaystyle \lim_{x \to 1} \dfrac{x - 1}{\sqrt{x + 3} - 2}
17. \displaystyle \lim_{x \to 2} \dfrac{\sqrt{3x - 2} - 2}{x^2 - 4}
18. \displaystyle \lim_{x \to 0} \dfrac{\sqrt{9 + x} - \sqrt{9 - x}}{x}
19. \displaystyle \lim_{h \to 0} \dfrac{\frac{1}{\sqrt{4 + h}} - \frac{1}{2}}{h}
20. \displaystyle \lim_{x \to 2} \dfrac{\sqrt{x + 7} - 1}{x - 2}

Reasoning

21. Wiebe says f(x) = \dfrac{x^2 - 2x}{x - 2} and g(x) = x are the same function. Explain what is wrong with the claim, and why the two still have the same limit as x \to 2.
22. Evaluate \lim_{x \to 1} \dfrac{x^3 - 3x + 2}{x^2 - 2x + 1}. After one factor x - 1 is canceled, what form does substitution give, and why does the factor theorem promise that another cancellation is possible?
23. Hylke says \displaystyle \frac{0}{0} must equal 1, since a number divided by itself is 1. Give four limits of the form \displaystyle \frac{0}{0} as x \to 0: one equal to 1, one equal to 5, one equal to 0, and one that does not exist.
24. Tjitske evaluates \displaystyle \lim_{x \to 1} \dfrac{\sqrt{x + 15} - 4}{x - 1} by multiplying top and bottom by \displaystyle \sqrt{x + 15} + 4. Explain why this cannot change the limit, why multiplying by \displaystyle \sqrt{x + 15} - 4 over itself would not help, and finish the limit.
25. Evaluate \displaystyle \lim_{x \to 8} \dfrac{\sqrt[3]{x} - 2}{x - 8} by writing x - 8 as a difference of cubes with \displaystyle a = \sqrt[3]{x} and b = 2.