12.2Finding Limits: Properties of Limits
Apply the sum, difference, product, quotient, and power limit laws to evaluate limits exactly. Use direct substitution, factoring, and rationalization to resolve indeterminate 0/0 forms.
Limit laws transform limit evaluation from estimation into exact computation. Direct substitution, factoring, and rationalizing are the algebraic toolkit every calculus student needs to evaluate derivatives and integrals confidently.
What Are Limit Laws?
Limit laws let you break a complicated limit into smaller, manageable pieces. Instead of building a table of values every time, you can apply algebraic rules to evaluate limits exactly — as long as the individual limits exist.
Vocabulary
Limit Law
A rule that lets you compute the limit of a combined expression using the limits of its parts.
Example: lim(x→2)[3x + 1] = 3·lim(x→2)[x] + lim(x→2)[1] = 6 + 1 = 7
Direct Substitution
Plugging the target value directly into the function. Works whenever the function is continuous at that point.
Example: lim(x→4) (x² + 1) = 4² + 1 = 17
Indeterminate Form (0/0)
When direct substitution gives 0/0, the limit may still exist — you must simplify first by factoring or rationalizing.
Example: lim(x→2) (x²−4)/(x−2) → 0/0 → factor → lim = 4
Squeeze Theorem
If g(x) ≤ f(x) ≤ h(x) near a, and lim g(x) = lim h(x) = L, then lim f(x) = L.
Example: −x² ≤ x²sin(1/x) ≤ x² near 0, so lim(x→0) x²sin(1/x) = 0
Worked Examples
Evaluate lim(x→3) [2x² + 5x − 1] using limit laws.
Apply the Sum and Constant Multiple laws to split the limit.
lim(x→3)[2x²] + lim(x→3)[5x] − lim(x→3)[1]
= 2·(lim x)² + 5·(lim x) − 1 [Power and Constant Multiple laws]
= 2·(3)² + 5·(3) − 1
= 2·9 + 15 − 1
= 18 + 15 − 1
Evaluate lim(x→−2) (x³ − 4x + 1) by direct substitution.
The function is a polynomial — continuous everywhere — so direct substitution works.
Substitute x = −2:
(−2)³ − 4(−2) + 1
= −8 + 8 + 1
Evaluate lim(x→2) (x² − 4)/(x − 2). Direct substitution gives 0/0 — resolve by factoring.
Direct substitution: (4 − 4)/(2 − 2) = 0/0 ← indeterminate form.
Factor the numerator: x² − 4 = (x − 2)(x + 2).
(x − 2)(x + 2)/(x − 2) — cancel (x − 2), valid since x ≠ 2.
Simplified expression: x + 2.
Now take the limit: lim(x→2) (x + 2) = 2 + 2.
Evaluate lim(x→9) (√x − 3)/(x − 9) by rationalizing.
Direct substitution: (√9 − 3)/(9 − 9) = 0/0 ← indeterminate form.
Multiply numerator and denominator by the conjugate (√x + 3):
[(√x − 3)(√x + 3)] / [(x − 9)(√x + 3)]
= (x − 9) / [(x − 9)(√x + 3)]
Cancel (x − 9): 1/(√x + 3).
lim(x→9) 1/(√x + 3) = 1/(√9 + 3) = 1/(3 + 3) = 1/6.
Use the Squeeze Theorem to evaluate lim(x→0) x²·sin(1/x).
sin(1/x) oscillates between −1 and 1 for all x ≠ 0.
So: −x² ≤ x²·sin(1/x) ≤ x² for all x ≠ 0.
lim(x→0) (−x²) = 0 and lim(x→0) x² = 0.
Both bounding limits equal 0.
By the Squeeze Theorem, the limit of the middle function must also equal 0.
Guided Practice
Evaluate lim(x→4) (3x² − 2x + 1) using direct substitution.
Hint: Polynomials are continuous everywhere — just substitute x = 4 directly.
Evaluate lim(x→5) (x² − 25)/(x − 5) by factoring.
Hint: Factor x² − 25 as (x − 5)(x + 5), then cancel the common factor.
Evaluate lim(x→0) (√(x + 4) − 2)/x by rationalizing.
Hint: Multiply top and bottom by the conjugate (√(x + 4) + 2) to eliminate the radical in the numerator.
Evaluate lim(x→−1) (x³ + 1)/(x + 1) by factoring.
Hint: Factor x³ + 1 using the sum of cubes formula: a³ + b³ = (a + b)(a² − ab + b²).
Given that −x² ≤ f(x) ≤ x² for all x near 0, find lim(x→0) f(x) using the Squeeze Theorem.
Hint: Compute lim(x→0)(−x²) and lim(x→0)(x²). If both equal the same value L, the Squeeze Theorem gives lim f(x) = L.
Common Mistakes
Common Mistakes
Stopping at 0/0 and writing "the limit does not exist."
0/0 is an indeterminate form — it means you must simplify further (factor or rationalize) before concluding.
Canceling (x − a) without noting the restriction x ≠ a.
You may cancel (x − a) because limits approach but never equal a, so x ≠ a is automatically satisfied.
Applying the Quotient Law when the denominator limit is 0.
The Quotient Law requires lim g(x) ≠ 0. If it is 0, use factoring or rationalizing first.
Forgetting to rationalize both numerator and denominator when multiplying by a conjugate.
Multiply the entire fraction (top AND bottom) by the conjugate to keep the value unchanged.
Math Tips
Math Tips
Always try direct substitution first — if the function is continuous at the point, you are done.
If you get 0/0, look for a common factor in the numerator and denominator to cancel.
If you see a radical in a difference, multiply by the conjugate to rationalize.
The Squeeze Theorem is your best tool when a function oscillates (like sin or cos) multiplied by something that goes to 0.
After simplifying, always re-substitute to confirm you no longer get an indeterminate form.
Interactive Practice — 5 Questions
Which limit law allows you to write lim(x→a)[f(x) + g(x)] = lim f(x) + lim g(x)?
Evaluate lim(x→3) (x² + 2x − 1) by direct substitution.
What is lim(x→3) (x² − 9)/(x − 3)?
To evaluate lim(x→4) (√x − 2)/(x − 4), the best first step is:
The Squeeze Theorem can be used to find lim(x→0) x²cos(1/x) because: