Source: Offline Homework 2.1, Section 2.1 | Calculus for Business, FIU
Tags: limits, limit notation, evaluating limits, direct substitution, factoring limits, rationalisation, limits at infinity, calculus for business, FIU, Section 2.1
Difficulty: Introductory to Intermediate
Prerequisites: Algebra fundamentals (factoring quadratics, simplifying rational expressions, working with radicals).
Big Picture
Limits are the foundation of calculus. Before you can touch derivatives or integrals, you need to understand what it means for a function to approach a value. This section covers the notation rules your professor expects and the core techniques for evaluating limits, both at finite points and at infinity. If you are coming into this cold, think of a limit as the answer to: "What value is this function heading towards as x gets closer and closer to some number?" Everything in Chapters 2 and 3 builds on this.
TL;DR
A limit describes the value a function approaches as x moves towards a given point (or towards infinity). You evaluate limits using direct substitution, algebraic simplification (factoring, rationalising), and leading-term analysis for limits at infinity. Correct notation matters for full marks: keep the "lim" symbol in every intermediate step until you have a plain number.
Limit
The value that f(x) approaches as x approaches a particular number (or infinity). You write it as lim(x→c) f(x) = L.
Think of it as: the y-value the function is heading towards, even if it never actually gets there.
Limit notation
The formal way of writing a limit statement: lim(x→c) must appear in front of every expression until you reach a final numerical answer. Dropping the "lim" symbol too early is the most common notation error on this homework.
In simple terms, this means: keep writing "lim" until there are no more x's left in the expression.
Direct substitution
Plugging the value of c directly into f(x) to evaluate lim(x→c) f(x). Works whenever the function is continuous at that point (no division by zero, no 0/0).
Think of it as: the first thing you try. If it gives you a real number, you are done.
Indeterminate form (0/0)
What you get when direct substitution produces 0 in both the numerator and denominator. This does not mean the limit is zero or undefined; it means you need to do more algebra (factor, rationalise, simplify) before evaluating.
In simple terms, this means: 0/0 is a signal to keep working, not an answer.
Polynomial function
A function built from non-negative integer powers of x, combined with addition, subtraction, and multiplication by constants. Examples: 3x⁴ − 5x + 7, or (3x − 1)(2x + 5) when expanded.
Think of it as: no fractions with x in the denominator, no roots, no absolute values.
Rational function
A ratio (fraction) of two polynomial functions. Examples: 3/(x − 1), or (x² − 2x − 1)/(x³ + 2).
In simple terms, this means: polynomial divided by polynomial.
One-sided limit
The limit of f(x) as x approaches c from one direction only. lim(x→c⁺) means from the right (values larger than c); lim(x→c⁻) means from the left (values smaller than c).
Think of it as: looking at the graph from only one side of the point.
The "lim" symbol must remain in front of any expression that still contains the variable x. Once you substitute and the expression is a pure number, the "lim" drops.
Correct notation pattern
lim(x→4) 2x² + 7 = lim(x→4) 2(16) + 7 = 39
The "lim(x→4)" stays through the substitution step. It disappears only at the final numerical answer. This is correct (C).
lim(x→3) (x+3)(x−5)/(x+3) = lim(x→3) (x − 5) = −2
The common factor (x+3) cancels first, then the limit is applied. The "lim" persists until the final step. Correct (C).
Incorrect notation patterns
lim(x→−1) 5/(x+3) = lim(x→−1) = 5/2
The second "lim(x→−1)" has an equals sign right after it with no expression attached. The "lim" should sit in front of the fraction, not in front of nothing. Corrected: lim(x→−1) 5/(x+3) = 5/2. Incorrect (IC).
lim(x→∞) (2x² − 3x)/(3x² − 7) = 2x²/3x² = 2/3
The "lim" was dropped while x² terms are still in the expression. After dividing out lower-order terms, the expression 2x²/3x² still contains x and needs the "lim" in front of it. Corrected: lim(x→∞) (2x² − 3x)/(3x² − 7) = lim(x→∞) 2x²/3x² = 2/3. Incorrect (IC).
The rule in one line: if x is still in the expression, "lim" stays. No exceptions.
Technique 1: Direct substitution
Always try this first. Plug x = c into the function. If you get a real number (not 0/0 or a/0), that number is the limit.
Example: lim(x→4) (−10) = −10
The function is a constant. No matter what x does, the output is −10.
Example: lim(x→−3) (x+4)/(2x−5) = (−3+4)/(2(−3)−5) = 1/(−11) = −1/11
Substituting x = −3 gives a defined value. Done.
Technique 2: Factor and cancel
When direct substitution gives 0/0, factor the numerator and denominator. Cancel the common factor, then substitute again.
Example: lim(x→2) (x²+2x−8)/(x²−4)
Direct substitution gives (4+4−8)/(4−4) = 0/0. Indeterminate.
Factor: (x+4)(x−2) / (x+2)(x−2)
Cancel (x−2): lim(x→2) (x+4)/(x+2)
Now substitute: (2+4)/(2+2) = 6/4 = 3/2
Technique 3: Rationalise (multiply by the conjugate)
Used when the expression involves a square root and direct substitution gives 0/0.
Example: lim(x→36) (x−36)/(√x−6)
Direct substitution gives 0/0.
Multiply numerator and denominator by (√x+6):
(x−36)(√x+6) / (√x−6)(√x+6) = (x−36)(√x+6) / (x−36)
Cancel (x−36): lim(x→36) (√x+6) = √36+6 = 6+6 = 12
Technique 4: Limits at infinity
For rational functions as x→∞ or x→−∞, compare the highest power in the numerator with the highest power in the denominator.
If the degrees are equal, the limit is the ratio of the leading coefficients.
If the numerator's degree is lower, the limit is 0.
If the numerator's degree is higher, the limit is ∞ or −∞.
Limits at infinity for non-rational expressions
For products or sums of polynomials, identify the highest-power term overall. That term dominates.
Example: lim(x→−∞) (x³−1)(2x²+5)
Expand the leading terms: x³ · 2x² = 2x⁵
As x→−∞, 2x⁵ → −∞ (odd power, so the sign flips)
Answer: −∞
Example: lim(x→∞) (3x⁴ − x⁵ + 2)
The highest-power term is −x⁵
As x→∞, −x⁵ → −∞
Answer: −∞
Difference of squares (used constantly in limit problems):
a² − b² = (a + b)(a − b)
Example: x² − 4 = (x + 2)(x − 2)
Factoring a quadratic (ax² + bx + c):
Find two numbers that multiply to give a·c and add to give b, then split the middle term.
Example: x² + 2x − 8 = (x + 4)(x − 2)
Conjugate multiplication (for roots):
Multiply by (√x + a)/(√x + a) to eliminate the square root in the denominator.
Key identity: (√x − a)(√x + a) = x − a²
Leading-term dominance (limits at infinity):
For lim(x→±∞) of any polynomial or rational function, only the highest-power term matters. Everything else becomes negligible.
Limits are how businesses model marginal cost and marginal revenue. When economists say "the cost of producing one more unit," they are describing a limit. The derivative, which you will meet in the next chapter, is defined as a limit, so every application of derivatives (profit maximisation, cost minimisation, demand elasticity) traces back to this concept.
In finance, limits at infinity describe long-run behaviour: what happens to average cost as production scales up, or what happens to an investment's value over an unbounded time horizon.
Students often think 0/0 means the limit does not exist. It does not. 0/0 is an indeterminate form, which means the limit might exist but you need to simplify first.
Students often drop the "lim" symbol as soon as they start substituting numbers. The "lim" must stay until every instance of x is gone from the expression.
Students often assume that a limit at infinity of a polynomial is always ∞. The sign of the leading term matters. If the leading coefficient is negative and the power is odd, the result as x→∞ is −∞.
Students sometimes confuse lim(x→c) f(x) with f(c). The limit describes where the function is heading; f(c) is the actual value at x = c. These can be different, especially when there is a hole in the graph.
⚠️ Notation is graded. Your professor explicitly states that proper limit notation is required for full credit. Do not skip the "lim" symbol in intermediate steps.
⚠️ The 0/0 indeterminate form appears in nearly every limit problem on this assignment. If you cannot factor or rationalise confidently, practise those algebra skills before the exam.
⚠️ Limits at infinity questions test whether you know which term dominates. For rational functions, compare degrees. For polynomials, identify the highest-power term and its sign.
⚠️ The difference between lim(x→c) f(x) and f(c) is likely to appear on the exam (see Question 3 on the homework). Know the distinction cold.
True or False: If direct substitution gives 0/0, the limit does not exist.
False. 0/0 is indeterminate; you need to simplify further.
Fill in the blank: The "lim" symbol should remain in the expression until ______.
All instances of the variable x have been replaced by numbers.
True or False: lim(x→∞) (5x³ − 2x⁴) = ∞.
False. The dominant term is −2x⁴, so the limit is −∞.
True or False: lim(x→c) f(x) always equals f(c).
False. The limit can exist even when f(c) is undefined, or when f(c) has a different value.
Fill in the blank: To evaluate lim(x→36) (x−36)/(√x−6), you multiply by the ______ of the denominator.
Conjugate. Multiply by (√x+6)/(√x+6).
Q: Evaluate lim(x→−3) (x+4)/(2x−5) using proper limit notation.
A: lim(x→−3) (x+4)/(2x−5) = (−3+4)/(2(−3)−5) = 1/(−11) = −1/11
Q: Evaluate lim(x→2) (x²+2x−8)/(x²−4). Show all steps.
A: Direct substitution gives 0/0. Factor: lim(x→2) (x+4)(x−2)/((x+2)(x−2)) = lim(x→2) (x+4)/(x+2) = 6/4 = 3/2
Q: Evaluate lim(x→36) (x−36)/(√x−6).
A: Direct substitution gives 0/0. Multiply by conjugate: lim(x→36) (x−36)(√x+6)/((√x−6)(√x+6)) = lim(x→36) (x−36)(√x+6)/(x−36) = lim(x→36) (√x+6) = 6+6 = 12
Q: Evaluate lim(x→−∞) (x³−1)(2x²+5).
A: The leading terms multiply to 2x⁵. As x→−∞, x⁵→−∞, so 2x⁵→−∞. The limit is −∞.
Q: Evaluate lim(x→∞) (3x⁴ − x⁵ + 2).
A: The dominant term is −x⁵. As x→∞, −x⁵→−∞. The limit is −∞.
Q: Is the following notation correct? lim(x→−1) 5/(x+3) = lim(x→−1) = 5/2. If not, why?
A: Incorrect. The second "lim" is followed by "=" with no expression. Correct version: lim(x→−1) 5/(x+3) = 5/2.
This material connects directly to Section 2.2 (continuity), where you will use limits to determine whether a function is continuous at a point. A function is continuous at x = c precisely when lim(x→c) f(x) = f(c). You have already seen this idea in Question 3 of the homework.
Limits also form the definition of the derivative in Chapter 3. The derivative is defined as lim(h→0) [f(x+h) − f(x)]/h, so every derivative you compute in this course is, at its core, a limit evaluation.
If you are comfortable with factoring, conjugate multiplication, and leading-term analysis, the derivative chapter will go more smoothly.
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