Algebraic Techniques for Limits

When you substitute into a limit and get $\tfrac{0}{0}$, the limit isn’t broken. It often means the top and bottom share a hidden factor that’s causing the trouble. Algebra (factoring, conjugates, common denominators) lets you remove that factor and then substitute. These techniques show up constantly, including on the no-calculator part of the AP exam and later in the definition of the derivative.

Key ideas

Why 0/0 needs more work

The form $\tfrac{0}{0}$ is indeterminate: on its own, it tells you nothing about the limit. For example,

$$\lim_{x \to 0} \frac{2x}{x} = 2, \qquad \lim_{x \to 0} \frac{x^2}{x} = 0, \qquad \lim_{x \to 0} \frac{x}{x^3} \ \text{does not exist}$$

All three give $\tfrac{0}{0}$ when you substitute. You have to rewrite the expression to see what’s really going on.

Why cancelling is allowed

$$\frac{x^2 - 9}{x - 3} = \frac{(x - 3)(x + 3)}{x - 3} = x + 3 \quad\text{for } x \ne 3$$

The two functions are equal everywhere except at $x = 3$. A limit never looks at $x = 3$ itself, so they have the same limit:

$$\lim_{x \to 3} \frac{x^2 - 9}{x - 3} = \lim_{x \to 3} (x + 3) = 6$$

The graph is the line $y = x + 3$ with a hole. The limit is the height of the hole.

The main techniques

When you see	Try
Polynomials on top and bottom	Factor both and cancel the common factor
A square root plus or minus a number	Multiply top and bottom by the conjugate, e.g. $\sqrt{x + 9} + 3$
Fractions inside a fraction	Combine them over a common denominator
Brackets with powers, like $(3 + h)^2$	Expand and simplify
Trig functions	Use an identity (like $\sin^2 x = 1 - \cos^2 x$) or the special trig limits

Choosing a procedure

1. Substitute first. If you get a real number, that’s the limit. Done.
2. Nonzero over zero (like $\tfrac{5}{0}$): the function blows up. The limit is $\infty$, $-\infty$, or does not exist. Check signs on each side (infinite limits).
3. Zero over zero: use algebra from the table above, then substitute again.
4. Piecewise or absolute value: find the one-sided limits separately.

Worked examples

Example 1: Factoring

Find $\displaystyle\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$.

Solution. Substituting gives $\tfrac{0}{0}$. Factor the difference of squares and cancel:

$$\lim_{x \to 3} \frac{x^2 - 9}{x - 3} = \lim_{x \to 3} \frac{(x - 3)(x + 3)}{x - 3} = \lim_{x \to 3} (x + 3) = 6$$

Example 2: Rationalizing with a conjugate

Find $\displaystyle\lim_{x \to 0} \frac{\sqrt{x + 9} - 3}{x}$.

Solution. Substituting gives $\tfrac{3 - 3}{0} = \tfrac{0}{0}$. Multiply top and bottom by the conjugate $\sqrt{x + 9} + 3$. The top becomes a difference of squares:

$$\begin{aligned} \lim_{x \to 0} \frac{\sqrt{x + 9} - 3}{x} \cdot \frac{\sqrt{x + 9} + 3}{\sqrt{x + 9} + 3} &= \lim_{x \to 0} \frac{(x + 9) - 9}{x\left(\sqrt{x + 9} + 3\right)} \\ &= \lim_{x \to 0} \frac{x}{x\left(\sqrt{x + 9} + 3\right)} \\ &= \lim_{x \to 0} \frac{1}{\sqrt{x + 9} + 3} = \frac{1}{3 + 3} = \frac{1}{6} \end{aligned}$$

Example 3: A common denominator

Find $\displaystyle\lim_{x \to 0} \frac{\dfrac{1}{x + 2} - \dfrac{1}{2}}{x}$.

Solution. Substituting gives $\tfrac{0}{0}$. Combine the fractions on top over the common denominator $2(x + 2)$:

$$\frac{1}{x + 2} - \frac{1}{2} = \frac{2 - (x + 2)}{2(x + 2)} = \frac{-x}{2(x + 2)}$$

Dividing by $x$ cancels the $x$:

$$\lim_{x \to 0} \frac{-x}{2x(x + 2)} = \lim_{x \to 0} \frac{-1}{2(x + 2)} = -\frac{1}{4}$$

Example 4: Choosing a procedure

Let $f(x) = \dfrac{x^2 + x - 6}{x^2 - 4}$. Find $\displaystyle\lim_{x \to 2} f(x)$ and $\displaystyle\lim_{x \to -2} f(x)$.

Solution. Factor once, then deal with each point:

$$f(x) = \frac{(x + 3)(x - 2)}{(x + 2)(x - 2)} = \frac{x + 3}{x + 2} \quad\text{for } x \ne 2$$

At $x = 2$, substituting into the original gives $\tfrac{0}{0}$, so use the simplified form:

$$\lim_{x \to 2} f(x) = \frac{2 + 3}{2 + 2} = \frac{5}{4}$$

At $x = -2$, substituting gives $\tfrac{-4}{0}$: nonzero over zero. In the simplified form, the top approaches $1$. Just right of $-2$, the bottom $x + 2$ is a small positive number; just left, a small negative number:

$$\lim_{x \to -2^+} f(x) = \infty, \qquad \lim_{x \to -2^-} f(x) = -\infty$$

So $\displaystyle\lim_{x \to -2} f(x)$ does not exist (there’s a vertical asymptote at $x = -2$).

Common mistakes

Writing ”= 0/0” as an answer. $\tfrac{0}{0}$ means “keep going”. On the AP exam, a final answer of $\tfrac{0}{0}$ or “undefined” earns nothing when the limit actually exists.

Dropping the limit symbol too early. Keep writing $\displaystyle\lim_{x \to a}$ on every line until you substitute. Writing $\tfrac{x^2 - 9}{x - 3} = x + 3 = 6$ says something false.

Cancelling terms instead of factors. In $\dfrac{x^2 + x - 6}{x^2 - 4}$, you can’t cancel the $x^2$‘s. Only cancel factors that multiply the whole top and the whole bottom.

Mishandling the conjugate. Multiply the top and the bottom by the conjugate. Leave the bottom factored; usually the troublesome factor cancels, and expanding just hides it.

Assuming nonzero/0 means the limit is infinity. The two sides can go in opposite directions, as in Example 4. Check the sign of the expression on each side.

Practice

1. (Warm-up) Find $\displaystyle\lim_{x \to 5} \frac{x^2 - 25}{x - 5}$.

Solution

$$\lim_{x \to 5} \frac{(x - 5)(x + 5)}{x - 5} = \lim_{x \to 5} (x + 5) = 10$$

2. (Warm-up) Find $\displaystyle\lim_{x \to -1} \frac{x^2 + 3x + 2}{x + 1}$.

Solution

$$\lim_{x \to -1} \frac{(x + 1)(x + 2)}{x + 1} = \lim_{x \to -1} (x + 2) = 1$$

3. (Warm-up) Find $\displaystyle\lim_{x \to 2} \frac{x^2 + 1}{x + 3}$. Do you need any algebra?

Solution

No. Substituting gives a real number right away:

$$\lim_{x \to 2} \frac{x^2 + 1}{x + 3} = \frac{5}{5} = 1$$

Always substitute first.

4. (Core) Find $\displaystyle\lim_{x \to 4} \frac{x - 4}{\sqrt{x} - 2}$.

Solution

Substituting gives $\tfrac{0}{0}$. Multiply by the conjugate $\sqrt{x} + 2$:

$$\lim_{x \to 4} \frac{(x - 4)(\sqrt{x} + 2)}{(\sqrt{x} - 2)(\sqrt{x} + 2)} = \lim_{x \to 4} \frac{(x - 4)(\sqrt{x} + 2)}{x - 4} = \lim_{x \to 4} (\sqrt{x} + 2) = 4$$

(Or factor $x - 4 = (\sqrt{x} - 2)(\sqrt{x} + 2)$ directly.)

5. (Core) Find $\displaystyle\lim_{x \to 1} \frac{x^3 - 1}{x^2 - 1}$.

Solution

Factor the difference of cubes and the difference of squares:

$$\lim_{x \to 1} \frac{(x - 1)(x^2 + x + 1)}{(x - 1)(x + 1)} = \lim_{x \to 1} \frac{x^2 + x + 1}{x + 1} = \frac{3}{2}$$

6. (Core) Find $\displaystyle\lim_{h \to 0} \frac{(3 + h)^2 - 9}{h}$.

Solution

Expand the top:

$$\lim_{h \to 0} \frac{9 + 6h + h^2 - 9}{h} = \lim_{h \to 0} \frac{h(6 + h)}{h} = \lim_{h \to 0} (6 + h) = 6$$

(You’ll see limits like this again: it’s the slope of $y = x^2$ at $x = 3$.)

7. (Core) Find $\displaystyle\lim_{x \to 3} \frac{\dfrac{1}{x} - \dfrac{1}{3}}{x - 3}$.

Solution

Combine the top over $3x$:

$$\frac{1}{x} - \frac{1}{3} = \frac{3 - x}{3x} = \frac{-(x - 3)}{3x}$$

So

$$\lim_{x \to 3} \frac{-(x - 3)}{3x(x - 3)} = \lim_{x \to 3} \frac{-1}{3x} = -\frac{1}{9}$$

8. (Challenge) Find $\displaystyle\lim_{x \to 2} \frac{x^2 - 4}{|x - 2|}$, if it exists.

Solution

Use one-sided limits. For $x \gt 2$, $|x - 2| = x - 2$:

$$\lim_{x \to 2^+} \frac{(x - 2)(x + 2)}{x - 2} = \lim_{x \to 2^+} (x + 2) = 4$$

For $x \lt 2$, $|x - 2| = -(x - 2)$:

$$\lim_{x \to 2^-} \frac{(x - 2)(x + 2)}{-(x - 2)} = \lim_{x \to 2^-} -(x + 2) = -4$$

The one-sided limits differ, so the limit does not exist.

9. (Challenge) Find $\displaystyle\lim_{x \to 0} \frac{\sin^2 x}{1 - \cos x}$. (Radians.)

Solution

Substituting gives $\tfrac{0}{0}$. Use $\sin^2 x = 1 - \cos^2 x = (1 - \cos x)(1 + \cos x)$:

$$\lim_{x \to 0} \frac{(1 - \cos x)(1 + \cos x)}{1 - \cos x} = \lim_{x \to 0} (1 + \cos x) = 1 + 1 = 2$$