Determining Limits Using Algebraic Properties of Limits: Study Guide (Calculus AB 1.5)

Course scope. This topic is shared by Calculus AB and Calculus BC. Everything here is examinable for both. It follows Topic 1.4, Estimating Limit Values from Tables.

A note on notation

This page has no equation renderer, so limits are written in a compact form:

• lim (x → a) f(x) means “the limit as x approaches a of f(x)”.
• lim (x → a⁻) f(x) is the left-hand limit: x approaches a through values less than a.
• lim (x → a⁺) f(x) is the right-hand limit: x approaches a through values greater than a.

On paper, write “x → a” under “lim” as usual. Keep “lim” on every line until you substitute.

Why we need properties of limits

In Topics 1.3 and 1.4 you estimated limits from graphs and tables. An estimate is useful, but it is not exact, and it is not a justification. This topic gives you rules that turn limits you already know into new limits. Each rule is a theorem, so each step you write can be justified by naming the rule.

Everything starts from two simple facts:

1. Constant: lim (x → a) c = c. A function that never changes approaches its own value.
2. Identity: lim (x → a) x = a. As x gets close to a, x gets close to a.

The properties of limits

Suppose lim (x → a) f(x) = L and lim (x → a) g(x) = M, where L and M are real numbers (both limits exist and are finite). Then:

Property	Rule	Condition
Sum	lim [f(x) + g(x)] = L + M	Both limits exist
Difference	lim [f(x) − g(x)] = L − M	Both limits exist
Constant multiple	lim [k · f(x)] = k · L	The limit of f exists; k is any constant
Product	lim [f(x) · g(x)] = L · M	Both limits exist
Quotient	lim [f(x) / g(x)] = L / M	Both limits exist and M ≠ 0
Power	lim [f(x)]ⁿ = Lⁿ	n is a positive whole number
Root	lim ⁿ√f(x) = ⁿ√L	For an even root (such as a square root), L > 0

All limits in the table are as x → a. Three things to notice:

• Every property needs the separate limits to exist. If lim f(x) does not exist, the sum property cannot be used. That does not prove the sum has no limit. It only means this rule cannot tell you.
• The quotient property fails when M = 0. If the top tends to a nonzero number and the bottom tends to 0, the quotient grows without bound near a, so there is no finite limit. If both tend to 0, the result 0/0 is undecided; Topic 1.6 deals with that.
• The same properties hold for one-sided limits. Replace “x → a” by “x → a⁻” or “x → a⁺” everywhere.

Why direct substitution works for polynomials

Take lim (x → 2) (3x² − 5x + 1). Using the properties one at a time:

• lim x = 2 (identity), so lim x² = 2² = 4 (power).
• lim 3x² = 3 × 4 = 12 and lim 5x = 5 × 2 = 10 (constant multiple).
• lim 1 = 1 (constant).
• So lim (3x² − 5x + 1) = 12 − 10 + 1 = 3 (sum and difference).

The result is the same as substituting x = 2. That is no accident. Every polynomial is built from constants and x by sums, differences, products and constant multiples, so for any polynomial p, lim (x → a) p(x) = p(a).

A rational function is a polynomial over a polynomial. By the quotient property, lim (x → a) p(x)/q(x) = p(a)/q(a), provided q(a) ≠ 0. For example,

lim (x → −1) (x² + 4)/(x − 2) = (1 + 4)/(−1 − 2) = 5/(−3) = −5/3

The bottom is −3, not 0, so substitution is justified.

Limits of composite functions

A composite function has an inner function inside an outer one, such as √(x³ + 1). The inner function is x³ + 1 and the outer function is the square root.

Composite property. If lim (x → a) g(x) = L, and the outer function f can be evaluated by substitution at L (that is, lim (y → L) f(y) = f(L)), then lim (x → a) f(g(x)) = f(L).

Polynomials, roots of positive numbers, sine and cosine all allow substitution at every point of their domain. So:

• lim (x → 2) √(x³ + 1): the inner limit is 8 + 1 = 9, and √ can be evaluated at 9. The limit is √9 = 3.
• lim (x → π) cos(x/3): the inner limit is π/3, and cosine can be evaluated anywhere. The limit is cos(π/3) = 1/2.

When the outer function jumps at L. If f has a jump or a hole at L, you cannot just write f(L). Instead ask: from which side does g(x) approach L? If g(x) approaches L from above (g(x) > L), the composite approaches lim (y → L⁺) f(y). If from below, use the left-hand limit of f. Worked example 2 part (f) shows this.

One-sided limits from formulas and graphs

From a formula. Use the piece of the formula that applies on that side. For example, |x| = −x when x < 0 and |x| = x when x > 0. So:

• lim (x → 0⁻) |x|/x = lim (x → 0⁻) (−x)/x = lim (x → 0⁻) (−1) = −1
• lim (x → 0⁺) |x|/x = lim (x → 0⁺) x/x = lim (x → 0⁺) 1 = 1

(Cancelling x is allowed because x ≠ 0 on either side.)

From a graph. Follow the curve towards x = a from one side and read the height it approaches. An open circle marks a point that is not on the graph; a filled circle marks the actual value f(a). The value f(a) never affects the limit.

The two-sided test. lim (x → a) f(x) exists exactly when both one-sided limits exist and are equal. Then the two-sided limit is that common value.

Worked example 1: combining known limits

Question. You are told that lim (x → 3) f(x) = 5 and lim (x → 3) g(x) = −2. Find each limit, or explain why the properties cannot give it.

(a) lim (x → 3) [f(x)g(x) + x²] / [f(x) − g(x)] (b) lim (x → 3) √(f(x) + 4) (c) lim (x → 3) [g(x)]³ (d) lim (x → 3) f(x) / [g(x) + 2]

(a) Work out the top and bottom separately first.

1. Top: lim f(x)g(x) = 5 × (−2) = −10 (product). lim x² = 3² = 9 (power of the identity). So the top tends to −10 + 9 = −1 (sum).
2. Bottom: lim [f(x) − g(x)] = 5 − (−2) = 7 (difference).
3. The bottom limit is 7, which is not 0, so the quotient property applies: the limit is −1/7.

(b) lim [f(x) + 4] = 5 + 4 = 9 (sum and constant). Since 9 > 0, the root property applies: the limit is √9 = 3.

(c) By the power property, the limit is (−2)³ = −8. Watch the sign: an odd power of a negative number is negative.

(d) The top tends to 5. The bottom tends to −2 + 2 = 0. The quotient property does not apply, because the bottom limit is 0. Since the top tends to a nonzero number while the bottom tends to 0, the quotient becomes unbounded near x = 3. There is no finite limit.

Check your reasoning. Each step names a property and confirms its condition (limits exist; bottom limit nonzero; positive number under a square root). That is what “justify” means in this topic.

Worked example 2: limits from graphs

Question. Use Figure 1 to find each limit, or explain why it does not exist.

(a) lim (x → 1) f(x) (b) lim (x → 1) g(x) (c) lim (x → 1) [f(x) + g(x)] (d) lim (x → 1) f(x)g(x) (e) lim (x → 3) f(g(x)) (f) lim (x → 1) f(g(x))

(a) From the left, the graph of f heads to the open circle at height 2. From the right, it heads to height 3. The one-sided limits are 2 and 3. They differ, so lim (x → 1) f(x) does not exist.

(b) From the left, g heads to 2. From the right, g heads to 1. The limit does not exist.

(c) The sum property cannot be used, because neither limit exists. Use one-sided limits instead; the properties do work on each side.

• Left: 2 + 2 = 4.
• Right: 3 + 1 = 4.

Both one-sided limits equal 4, so lim (x → 1) [f(x) + g(x)] = 4. The jumps cancel: f jumps up by 1 and g jumps down by 1.

(d) Left: 2 × 2 = 4. Right: 3 × 1 = 3. They differ, so the limit does not exist.

(e) Near x = 3, g(x) = x, so g(x) → 3. The graph of f has no break at 3, and f(3) = 1. By the composite property, lim (x → 3) f(g(x)) = 1.

(f) The inner limit does not exist at 1, so track each side.

• As x → 1⁺, g(x) = x approaches 1 from above. So f(g(x)) approaches the right-hand limit of f at 1, which is 3.
• As x → 1⁻, g(x) = 3 − x is a little more than 2 (for example, g(0.9) = 2.1). So g(x) approaches 2 from above. f has no break at 2 and f(2) = 2, so f(g(x)) → 2.

The one-sided limits are 3 and 2, so lim (x → 1) f(g(x)) does not exist.

Interpretation. Part (c) shows that a sum can have a limit even when its parts do not. Part (f) shows why the side of approach matters for composites.

Worked example 3: a piecewise function with a parameter

Question. Let h(x) = x² + kx for x < 2, h(2) = 7, and h(x) = 3x − 1 for x > 2, where k is a constant.

(a) Find lim (x → 2⁻) h(x) and lim (x → 2⁺) h(x). (b) Find the value of k for which lim (x → 2) h(x) exists, and give the limit. (c) For that value of k, find lim (x → 2) √(h(x) + 4).

(a) For x < 2, h(x) = x² + kx, a polynomial in x. So lim (x → 2⁻) h(x) = 2² + 2k = 4 + 2k. For x > 2, h(x) = 3x − 1, so lim (x → 2⁺) h(x) = 6 − 1 = 5.

(b) The two-sided limit exists only if 4 + 2k = 5. So 2k = 1 and k = 1/2. The limit is then 5. Notice that h(2) = 7 plays no part: the limit is 5 even though the function value is 7.

(c) The inner limit is 5, so h(x) + 4 → 9 > 0. The square root can be evaluated at 9, so the limit is √9 = 3 (composite property).

Check. With k = 1/2, h(1.99) = 1.99² + 0.995 = 4.9551, and h(2.01) = 5.03. Both are close to 5.

Common misconceptions

• “The limit properties always work.” Each needs the separate limits to exist, and the quotient property also needs a nonzero bottom limit. Check the conditions before you use a rule.
• “If lim f(x) does not exist, then lim [f(x) + g(x)] cannot exist.” False. Worked example 2(c) has a sum with a limit, built from two parts without limits.
• “A zero bottom limit means the limit is 0.” No. Nonzero over 0 means there is no finite limit. 0 over 0 is undecided (Topic 1.6).
• “The limit equals the function value.” Worked example 3 has limit 5 and value 7. A limit depends only on values near a, never at a.
• “lim f(g(x)) is always f(lim g(x)).” Only when f can be evaluated by substitution at the inner limit. If f jumps there, look at which side g approaches from.
• Sign slips with powers. (−2)³ = −8, but (−2)² = 4. Keep brackets around negative limits.
• Reading the filled dot as the limit. On a graph, the filled dot is f(a). The limit is where the curve heads, which may be an open circle.

Where this leads

Topic 1.6, Determining Limits Using Algebraic Manipulation, handles the case the quotient property cannot: a bottom limit of 0, especially 0/0. The one-sided reasoning here returns in Topics 1.10 to 1.12, where continuity is defined as “the limit equals the function value”, and the composite property becomes a statement about continuous functions. Students in either course can return to the Calculus AB roadmap or the Calculus BC roadmap.

Try the practice questions now, then use the revision notes and the checklist to consolidate.