DWG NO. 10.2 — Lesson 2 of 5

Evaluating Limits Algebraically

Unit 10: Limits and an Introduction to Calculus · ~25 min

Objective Evaluate limits algebraically using direct substitution, the limit laws, and factoring or rationalizing to resolve 0/0 forms.

Lesson 10.1 estimated limits from tables — useful for building intuition, but too slow for everyday use. Most limits you'll meet can be computed exactly with a few algebraic techniques, starting with the simplest one of all: just plugging in.

Direct substitution

If f is a polynomial, or a rational function whose denominator doesn't vanish at a, then f is continuous at a (Lesson 10.4 makes this precise), and the limit is simply the function value:

Direct substitution li\(m_{x→a}\) f(x) = f(a), whenever f(a) is defined by ordinary evaluation

The limit laws

Assuming li\(m_{x→a}\) f(x) and li\(m_{x→a}\) g(x) both exist, limits distribute over the usual operations:

Sum/difference: lim [f(x) ± g(x)] = lim f(x) ± lim g(x)
Product: lim [f(x)·g(x)] = lim f(x) · lim g(x)
Quotient: lim [f(x)/g(x)] = lim f(x) / lim g(x), provided lim g(x) ≠ 0
Power: lim [f(x)]ⁿ = [lim f(x)]ⁿ

These are why direct substitution works for polynomials and most rational functions in the first place — a polynomial is just sums, differences, and products of x and constants, and each piece substitutes cleanly.

The indeterminate form 0/0

Direct substitution breaks down when it produces 0/0 — that's not "the answer is 0," it's a signal that the expression needs more work before the limit can be read off. Two standard fixes:

Factor and cancel: if both numerator and denominator share a factor that vanishes at a, factor it out and cancel before substituting.
Rationalize: if a radical is causing the 0/0, multiply numerator and denominator by the conjugate to clear the radical, then cancel.

(x²−1)/(x−1) factors to (x+1)(x−1)/(x−1) = x+1 for x≠1 — canceling reveals the line, and the hole confirms li\(m_{x→1}\) = 2

Worked Example 1 · Direct substitution

ProblemEvaluate li\(m_{x→3}\) (2x² − 5x + 1).

1This is a polynomial, so substitute x = 3 directly: 2(3)² − 5(3) + 1.

22(9) − 15 + 1 = 18 − 15 + 1.

li\(m_{x→3}\) (2x²−5x+1) = 4

Worked Example 2 · Factor and cancel

ProblemEvaluate li\(m_{x→-2}\) (x²−4)/(x+2).

1Direct substitution gives 0/0 — an indeterminate form, so factor first: (x²−4) = (x−2)(x+2).

2(x−2)(x+2)/(x+2) = x−2 for all x ≠ −2, which is all that matters for the limit.

3Now substitute: li\(m_{x→-2}\) (x−2) = −2−2.

li\(m_{x→-2}\) (x²−4)/(x+2) = −4

Worked Example 3 · Rationalizing

ProblemEvaluate li\(m_{x→0}\) (\(\sqrt{x+4}\) − 2)/x.

1Direct substitution gives 0/0. Multiply numerator and denominator by the conjugate \(\sqrt{x+4}\) + 2.

2Numerator becomes (x+4) − 4 = x, so the whole expression is x / [x(\(\sqrt{x+4}\)+2)] = 1/(\(\sqrt{x+4}\)+2) for x≠0.

3Substitute x = 0: 1/(\(\sqrt{4}\) + 2) = 1/(2+2).

li\(m_{x→0}\) (\(\sqrt{x+4}\)−2)/x = 1/4

Guided practice

1.Evaluate li\(m_{x→1}\) (3x² − 2x + 5) by direct substitution.
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3(1)²−2(1)+5 = 3−2+5 = 6.
2.Evaluate li\(m_{x→5}\) (x²−25)/(x−5).
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Factor: (x−5)(x+5)/(x−5) = x+5. Substitute x=5: 10.
3.Evaluate li\(m_{x→-3}\) (x²+5x+6)/(x+3).
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Factor: (x+2)(x+3)/(x+3) = x+2. Substitute x=−3: −1.
4.Evaluate li\(m_{x→0}\) (\(\sqrt{x+9}\) − 3)/x.
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Rationalize: multiply by \(\sqrt{x+9}\)+3 to get x/[x(\(\sqrt{x+9}\)+3)] = 1/(\(\sqrt{x+9}\)+3). Substitute x=0: 1/(3+3) = 1/6.
5.Evaluate li\(m_{x→2}\) (x³−8)/(x−2).
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Factor as a difference of cubes: (x−2)(x²+2x+4)/(x−2) = x²+2x+4. Substitute x=2: 4+4+4 = 12.

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