Guides / Limits

Evaluating limits algebraically: plug in, factor, or use the conjugate

Most limits on a test are solved in one of three ways, and substitution always comes first. When it gives 0/0, that isn't the answer. It's the signal to factor, multiply by the conjugate, or (at infinity) divide by the highest power.

Short answer

Substitute the value first. If you get a number, that's the limit. If you get 0/0, the expression has to be rewritten: factor the top and bottom and cancel the common factor, or, if there's a square root, multiply top and bottom by the conjugate. Then substitute again. For a limit as x → ∞, divide every term by the highest power of x in the denominator: same degree on top and bottom gives the ratio of the leading coefficients, a bigger bottom gives 0.

Why it works

A limit asks what value f(x) gets close to as x gets close to a number, never what happens at that number. So two expressions that are equal for every x except the one you're approaching have the same limit. That's exactly what canceling does: (x − 3)(x + 2)x − 3 and x + 2 agree everywhere except at x = 3, where the first one is undefined. The limit never looks at x = 3, so you can use the simpler one.

0/0 comes up because the top and bottom share a factor that is zero at that point. Finding the factor, by factoring or by rationalizing with the conjugate, and canceling it is the whole job. It's the same canceling as in simplifying rational expressions.

Example 1: factor and cancel

Find lim_x→3 x² − x − 6x − 3.

9 − 3 − 63 − 3 = 00Substitute first. 0/0 means the top and bottom share a factor of (x − 3). Keep going.
lim_x→3 (x − 3)(x + 2)x − 3Factor the top: −3 and 2 multiply to −6 and add to −1.
lim_x→3 (x + 2)Cancel the common factor. Legal, because the limit never uses x = 3 itself.
5Now substitute: 3 + 2 = 5. Check with a nearby value: at x = 3.01 the original is 5.01.

Example 2: a square root, so use the conjugate

Find lim_x→9 √x − 3x − 9.

3 − 39 − 9 = 00Substitute first. 0/0 again, and there's nothing to factor on top.
√x − 3x − 9 · √x + 3√x + 3The conjugate of √x − 3 is √x + 3. Multiply the top and the bottom, so you're multiplying by 1.
x − 9(x − 9)(√x + 3)Top: (√x − 3)(√x + 3) = x − 9. Leave the bottom factored so you can see what cancels.
lim_x→9 1√x + 3Cancel (x − 9).
16Substitute: 1/(3 + 3). At x = 9.01 the original is about 0.16662, close to 1/6 ≈ 0.16667.

Example 3: a limit at infinity

Find lim_x→∞ 4x² − x2x² + 5.

lim_x→∞ 4 − 1x2 + 5x²Divide every term, top and bottom, by x², the highest power in the denominator.
4 − 02 + 0A fixed number divided by a bigger and bigger x goes to 0.
2Same degree on top and bottom, so the limit is the ratio of the leading coefficients, 4/2.

The shortcut for a fraction of polynomials as x → ∞: if the top and bottom have the same degree, the limit is the ratio of the leading coefficients. If the bottom's degree is bigger, the limit is 0. If the top's degree is bigger, the fraction grows without bound and there's no finite limit.

Common mistakes, and the exact line they happen on

1. Stopping at 0/0

Find lim_x→3 x² − x − 6x − 3 (Example 1).

00Right so far: this is what substitution gives.
The limit does not existThis is the mistake. 0/0 says the function isn't defined at 3, but the limit is about values near 3, and they all get close to 5. Writing 0, 1, or "undefined" is the same mistake.

Treat 0/0 as an instruction, not an answer: something cancels, so go find it. A limit genuinely doesn't exist when the two sides approach different values, or the function grows without bound. 0/0 on its own tells you neither.

2. Canceling terms instead of factors

Find lim_x→2 x² − 4x − 2.

x² − 4x − 2 = x − 4−2This is the mistake. One x was crossed off the top and the bottom. But x isn't a factor of the whole top or the whole bottom; it's part of a sum. This gives (2 − 4)/(−2) = 1.
(x − 2)(x + 2)x − 2 = x + 2Correct line: factor the difference of squares, then cancel (x − 2).
4Substitute: 2 + 2.

3. Multiplying only the top by the conjugate

Find lim_x→9 √x − 3x − 9 (Example 2).

(√x − 3)(√x + 3)x − 9 = x − 9x − 9 = 1This is the mistake. Only the top was multiplied by √x + 3, which multiplies the whole fraction by about 6 and changes its value. The bottom has to become (x − 9)(√x + 3). The answer is 1/6, not 1.

4. Using the leading coefficients when the degrees differ

Find lim_x→∞ 3x + 1x² + 2.

lim_x→∞ 3x + 1x² + 2 = 31 = 3This is the mistake. The ratio-of-leading-coefficients shortcut only applies when the top and bottom have the same degree. Here the top is degree 1 and the bottom is degree 2. At x = 1000 the fraction is about 0.003.
lim_x→∞ 3x + 1x²1 + 2x²Correct line: divide every term by x².
0The top goes to 0 and the bottom goes to 1.

Practice

Find lim_x→2 x² + 1x + 3.
Show answer
Substitute: (4 + 1)/(2 + 3) = 1. No 0/0, so there's nothing to rewrite. The limit is 1.
Find lim_x→−4 x² + 3x − 4x + 4.
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Substituting gives 0/0. Factor: (x + 4)(x − 1)/(x + 4) = x − 1, so the limit is −4 − 1 = −5.
Find lim_x→5 x² − 25x² − 3x − 10.
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(x − 5)(x + 5) / ((x − 5)(x + 2)) = (x + 5)/(x + 2). Substitute: 10/7.
Find lim_x→0 √x + 9 − 3x.
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Multiply top and bottom by √x + 9 + 3: the top becomes x, and the fraction becomes 1/(√x + 9 + 3). Substitute: 1/6.
Find lim_x→∞ 5x³ − 2x2x³ + x² + 1.
Show answer
Same degree (3) on top and bottom, so the limit is the ratio of leading coefficients: 5/2. Dividing every term by x³ gives (5 − 0)/(2 + 0 + 0).