Source: Offline Homework 2.1, Section 2.1 (Questions 3-6) | Calculus for Business, FIU
Tags: graphical limits, one-sided limits, piecewise functions, polynomial function, rational function, continuity, holes, calculus for business, FIU, Section 2.1
Difficulty: Introductory to Intermediate
Prerequisites: Familiarity with limit notation and basic limit evaluation (see companion notes: Limit Notation and Evaluation). Ability to read a graph and identify coordinates.
Big Picture
This section moves beyond algebraic limit evaluation into three connected areas: reading limits off a graph, handling functions defined in pieces, and classifying functions by type. These skills are tested differently from the algebra-based limit problems. Graph questions require you to trace the curve with your eye. Piecewise questions require you to pick the right formula for the right region. Classification questions test whether you understand what makes a function polynomial, rational, or neither. All three appear regularly on exams.
TL;DR
One-sided limits ask what happens as you approach a point from just the left or just the right. Piecewise functions use different formulas for different ranges of x, so you need to match the formula to the direction of approach. Function classification depends on whether the expression is built from polynomial operations alone, is a ratio of polynomials, or involves roots or other non-polynomial elements.
One-sided limit (left-hand limit)
lim(x→c⁻) f(x). The value f(x) approaches as x comes from values less than c (from the left on the number line).
Think of it as: sliding your finger along the curve from the left side and seeing where it ends up.
One-sided limit (right-hand limit)
lim(x→c⁺) f(x). The value f(x) approaches as x comes from values greater than c (from the right).
Think of it as: sliding your finger along the curve from the right side.
Two-sided limit
lim(x→c) f(x). This exists only when the left-hand and right-hand limits are equal. If they disagree, the two-sided limit does not exist (DNE).
In simple terms, this means: both sides have to agree, or there is no limit.
Piecewise function
A function defined by different formulas on different intervals of x. You pick which formula to use based on where x falls.
Think of it as: a set of rules with conditions. "If x < 4, use this formula. If x = 4, use that one. If x > 4, use the other."
Hole (removable discontinuity)
A point where the limit exists but the function either is not defined or has a different value. On a graph, it appears as an open circle.
In simple terms, this means: the curve wants to pass through that point, but something is missing or misplaced there.
Continuous at a point
A function is continuous at x = c when three conditions hold: f(c) is defined, lim(x→c) f(x) exists, and lim(x→c) f(x) = f(c).
Think of it as: you can draw the graph through x = c without lifting your pen.
When a problem gives you a graph and asks for a limit, you are not computing anything algebraically. You are reading the y-value the curve approaches.
Right-hand limit: lim(x→c⁺) f(x)
Start slightly to the right of x = c and trace the curve leftward towards c.
The y-value you approach is the right-hand limit.
Open circles (hollow dots) mean the function is not defined at that point, but the limit still exists if the curve approaches a value.
Left-hand limit: lim(x→c⁻) f(x)
Start slightly to the left of x = c and trace the curve rightward towards c.
The y-value you approach is the left-hand limit.
Two-sided limit: lim(x→c) f(x)
Exists only if the left-hand limit and the right-hand limit are the same number.
If they differ, write DNE (does not exist).
Function value: f(c)
This is the actual y-coordinate of the point at x = c.
Look for a solid (filled) dot on the graph at x = c. That dot's y-value is f(c).
If there is only an open circle at x = c and no solid dot, then f(c) is undefined.
f(c) can be different from the limit. This is the key distinction.
Example from the homework (Question 4):
The graph shows a curve with various behaviours at x = −2, x = 0, x = 2, and x = 6. For each, you trace the curve from the correct direction and read the y-value. The filled dot tells you f(c); the direction the curve approaches tells you the limit.
A piecewise function has different rules for different intervals. To find a limit, you must choose the correct rule based on the direction of approach.
The homework's piecewise function:
f(x) = 12 when x < 4, f(x) = 4 − x² when x = 4, f(x) = 3x + 1 when x > 4
Finding one-sided limits
lim(x→4⁺) f(x): x is approaching 4 from the right, so x > 4. Use the rule f(x) = 3x + 1.
lim(x→4⁺) (3x + 1) = 3(4) + 1 = 13
lim(x→4⁻) f(x): x is approaching 4 from the left, so x < 4. Use the rule f(x) = 12.
lim(x→4⁻) 12 = 12 (a constant)
Finding the two-sided limit
lim(x→4) f(x): the left-hand limit is 12 and the right-hand limit is 13. These are not equal, so the two-sided limit does not exist (DNE).
Finding f(4)
At x = 4 exactly, use the rule f(x) = 4 − x². So f(4) = 4 − 16 = −12.
Note that f(4) = −12 is different from both one-sided limits. The function's value at a point does not need to match the limit.
Limit at a point away from the boundary
lim(x→−2) f(x): since −2 < 4, use the rule f(x) = 12. The function is constant at 12 for all x < 4, so lim(x→−2) f(x) = 12.
Question 6 asks you to label each function as polynomial (P), rational (R), or neither (N). The rules are straightforward.
Polynomial (P)
Every term is a constant times x raised to a non-negative whole number. No x in a denominator, no roots of x, no absolute values.
f(x) = 3x⁴ − 5x + 7: Polynomial (P). All terms are non-negative integer powers of x.
f(x) = (3x − 1)(2x + 5): Polynomial (P). When expanded, this is 6x² + 13x − 5, all non-negative integer powers.
Rational (R)
A fraction where both the numerator and the denominator are polynomials.
g(x) = 3/(x − 1): Rational (R). The numerator is 3 (a polynomial) and the denominator is x − 1 (a polynomial).
m(x) = (x² − 2x − 1)/(x³ + 2): Rational (R). Both top and bottom are polynomials.
Neither (N)
Anything involving roots of x, absolute values of x, or other non-polynomial operations, even if it is written as a fraction.
h(x) = ∛x − 1/x + 1: Neither (N). The cube root of x is not a non-negative integer power, and 1/x is x to the power of −1.
f(x) = (√x − 4)/(x + 1): Neither (N). The numerator contains √x, which is x to the power of 1/2. That makes the numerator not a polynomial, so the overall function is not rational either.
Question 3 on the homework asks you to state two properties of f(x) = (x² + 2x − 8)/(x² − 4) at x = 2.
Property 1: The limit exists at x = 2.
From the evaluation in the companion notes, lim(x→2) f(x) = 3/2. Even though the function is not defined at x = 2 (you get 0/0 on substitution), the limit still has a value after simplification.
Property 2: The function is not defined at x = 2.
Substituting x = 2 directly gives 0/0, which means f(2) does not produce a real number. The function has a hole (removable discontinuity) at x = 2.
Taken together, these two properties tell you the function is not continuous at x = 2. The limit exists, but f(2) is not defined, so the continuity condition fails.
Students often think that if the graph has an open circle at a point, the limit does not exist there. The limit can still exist; the open circle only means the function is not defined (or has a different value) at that exact point.
Students sometimes use the formula for x = c to evaluate a one-sided limit approaching c. For a piecewise function, the formula at x = c applies only to f(c), not to the limits from the left or right. The limits use the formulas for x < c and x > c.
Students often classify any function with a fraction as "rational." A rational function specifically requires both the numerator and denominator to be polynomials. If either contains a root, an absolute value, or a non-integer power, the function is neither polynomial nor rational.
Students sometimes assume that if the two one-sided limits exist, the two-sided limit must also exist. The two-sided limit exists only when the two one-sided limits are equal.
⚠️ Graph-reading questions are easy marks if you are careful. Trace the curve from the correct direction. Do not confuse an open circle (limit may exist) with a filled dot (function value).
⚠️ Piecewise limit questions test whether you can match the correct formula to the correct direction (left or right). A common exam trap is giving x = c its own formula that differs from both one-sided limits.
⚠️ The "two properties" style question (Question 3) tests your understanding of continuity conditions. Know the three requirements for continuity and be able to identify which one fails.
⚠️ Function classification (P, R, N) is often a quick section of the exam. The key test: does every term have a non-negative integer power of x? If yes, polynomial. If it is a ratio of polynomials, rational. Anything else, neither.
True or False: If lim(x→3⁻) f(x) = 5 and lim(x→3⁺) f(x) = 7, then lim(x→3) f(x) = 6.
False. The two-sided limit does not exist because the one-sided limits are not equal. You do not average them.
Fill in the blank: For a piecewise function, to find lim(x→4⁺) f(x), use the formula for x ______ 4.
Greater than (>). You are approaching from the right, so x > 4.
True or False: f(x) = (√x − 4)/(x + 1) is a rational function.
False. The numerator contains √x, which is not a polynomial term. The function is neither polynomial nor rational.
True or False: If there is an open circle at (2, 5) on the graph, then f(2) = 5.
False. An open circle means the function does not take that value at that point. f(2) is either undefined or has a different value shown by a filled dot.
Fill in the blank: A function is continuous at x = c if f(c) is defined, lim(x→c) f(x) exists, and ______.
lim(x→c) f(x) = f(c). The limit and the function value must be the same.
Q: For the piecewise function f(x) = {12 if x < 4, 4 − x² if x = 4, 3x + 1 if x > 4}, find lim(x→4⁺) f(x).
A: Since x→4⁺ means x > 4, use f(x) = 3x + 1. lim(x→4⁺) (3(4) + 1) = 13.
Q: Does lim(x→4) f(x) exist for the piecewise function above? Why or why not?
A: No. The left-hand limit is 12 and the right-hand limit is 13. Since they are not equal, the two-sided limit does not exist.
Q: What is f(4) for the piecewise function above?
A: At x = 4, use the rule f(x) = 4 − x². f(4) = 4 − 16 = −12.
Q: Is h(x) = ∛x − 1/x + 1 a polynomial, rational, or neither?
A: Neither. The cube root is x to the power of 1/3, which is not a non-negative integer, and 1/x is x to the power of −1.
Q: State two properties of f(x) = (x² + 2x − 8)/(x² − 4) at x = 2.
A: (1) The limit exists: lim(x→2) f(x) = 3/2. (2) The function is not defined at x = 2 (direct substitution gives 0/0). Therefore f has a hole (removable discontinuity) at x = 2.
Q: If a graph shows an open circle at (2, 5) and a filled dot at (2, 3), what is lim(x→2) f(x) and what is f(2)?
A: It depends on the curve, but if the curve approaches y = 5 from both sides, then lim(x→2) f(x) = 5 and f(2) = 3. The limit and the function value are different.
The distinction between f(c) and lim(x→c) f(x) leads directly into the formal definition of continuity in Section 2.2. If you are comfortable with the three conditions for continuity, that section will feel like a short extension of what you already know.
Piecewise functions reappear in later chapters when modelling real-world situations where the rule changes at a threshold: tax brackets, shipping cost tiers, or tiered pricing. The skill of choosing the right formula for the right interval carries forward.
Function classification (polynomial, rational, neither) matters because each type has different rules for limits, continuity, and derivatives. Polynomials are continuous everywhere. Rational functions are continuous except where the denominator is zero. Functions that are "neither" need to be checked case by case.
Related Terms / Search Tags
graphical limits, one-sided limits, left-hand limit, right-hand limit, two-sided limit, piecewise function, piecewise-defined function, function value vs limit, open circle, filled dot, hole, removable discontinuity, continuous function, continuity conditions, polynomial function, rational function, function classification, calculus for business, FIU, Section 2.1