Limits and Continuity — Precalculus Preview
The complete guide to limits and continuity for precalculus and intro calculus. Covers the intuitive definition, one-sided limits, limit laws, indeterminate forms, the Squeeze Theorem, continuity conditions, types of discontinuity, the Intermediate Value Theorem, and the epsilon-delta definition — plus a preview of how limits lead directly to derivatives.
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1. The Intuitive Idea of a Limit
Imagine tracing the graph of a function y = f(x) with your finger from both sides toward the point where x equals some number a. The value your finger approaches — but does not necessarily touch — is the limit of f(x) as x approaches a.
The core insight: the limit does not care what happens exactly at x equals a. The function may be undefined there (a hole in the graph), or defined with a different value. What matters is the behavior near a, not at a.
Limit Notation
“As x approaches a, f(x) approaches L”
Worked Example 1.1
Estimate the limit as x approaches 2 of f(x) = (x squared minus 4) divided by (x minus 2) by building a table of values.
| x | f(x) = (x^2 - 4) / (x - 2) |
|---|---|
| 1.9 | 3.9 |
| 1.99 | 3.99 |
| 1.999 | 3.999 |
| 2.001 | 4.001 |
| 2.01 | 4.01 |
| 2.1 | 4.1 |
The table suggests: as x approaches 2 from either side, f(x) approaches 4. Therefore the limit as x approaches 2 of f(x) = 4. Note that f(2) is undefined (division by zero), yet the limit exists.
Key Distinction
The limit of f(x) as x approaches a is NOT the same as f(a). The limit asks: “what value does f(x) get close to?” The function value asks: “what does f actually equal at a?” These can differ — or one may not exist when the other does.
2. One-Sided Limits
Sometimes a function approaches different values depending on which side of a you approach from. One-sided limits capture each direction separately.
Left-Hand Limit
x approaches a from the left (x is less than a). The superscript minus indicates approach from below.
Right-Hand Limit
x approaches a from the right (x is greater than a). The superscript plus indicates approach from above.
Worked Example 2.1 — Piecewise Function
Let f(x) = x plus 1 when x is less than 3, and f(x) = (x minus 3) squared plus 2 when x is greater than or equal to 3. Find both one-sided limits as x approaches 3.
Left limit: use the piece x plus 1. As x approaches 3 from the left: 3 plus 1 = 4. So the left-hand limit = 4.
Right limit: use the piece (x minus 3) squared plus 2. As x approaches 3 from the right: 0 plus 2 = 2. So the right-hand limit = 2.
Since 4 ≠ 2, the two-sided limit does not exist at x = 3 (jump discontinuity).
Worked Example 2.2 — Absolute Value
Find the one-sided limits of f(x) = |x| divided by x as x approaches 0.
For x greater than 0: |x| divided by x = x divided by x = 1. Right-hand limit = 1.
For x less than 0: |x| divided by x = negative x divided by x = -1. Left-hand limit = -1.
Since 1 ≠ -1, the two-sided limit does not exist. This is also a jump discontinuity.
3. Limit Existence Condition
Theorem
The two-sided limit of f(x) as x approaches a exists and equals L if and only if both one-sided limits exist and equal L:
The three situations where a limit does not exist (DNE):
- 1.The left-hand and right-hand limits are different (jump discontinuity as in the piecewise example above).
- 2.The function oscillates infinitely many times near x equals a, so it never settles on a single approach value. Classic example: sin(1/x) as x approaches 0.
- 3.The function grows without bound near x equals a (goes to positive or negative infinity). The limit is said to be infinite, which technically means DNE as a real number.
4. Limits at Infinity — Horizontal Asymptotes
When x grows without bound (either toward positive infinity or negative infinity), the limit describes the function's end behavior. If the limit equals a finite number L, then y = L is a horizontal asymptote of the graph.
Technique: Divide by the Highest Power of x
For rational functions, divide every term in the numerator and denominator by the highest power of x present. Then apply: as x approaches infinity, 1 divided by x to any positive power approaches 0.
Worked Example 4.1 — Degree Equal
Find the limit as x approaches positive infinity of (3x squared plus 2x) divided by (x squared minus 5).
Divide numerator and denominator by x squared:
(3 + 2/x) / (1 - 5/x squared)
As x → infinity: 2/x → 0 and 5/x² → 0
Limit = 3/1 = 3. Horizontal asymptote: y = 3
Worked Example 4.2 — Numerator Degree Higher
Find the limit as x approaches infinity of (x cubed plus 1) divided by (x squared plus 4).
Divide by x squared:
(x + 1/x squared) / (1 + 4/x squared)
As x → infinity: numerator → infinity, denominator → 1
Limit = infinity. No horizontal asymptote.
Worked Example 4.3 — Denominator Degree Higher
Find the limit as x approaches infinity of (2x plus 7) divided by (x squared plus 1).
Divide by x squared:
(2/x + 7/x squared) / (1 + 1/x squared)
As x → infinity: numerator → 0, denominator → 1
Limit = 0. Horizontal asymptote: y = 0
Degree Rule Summary
- Degree of numerator equals degree of denominator: limit = ratio of leading coefficients
- Degree of numerator is greater: limit = plus or minus infinity (no H.A.)
- Degree of denominator is greater: limit = 0 (H.A. at y = 0)
5. Infinite Limits — Vertical Asymptotes
An infinite limit occurs when a function grows without bound as x approaches a finite value a. This corresponds to a vertical asymptote at x = a.
f(x) grows toward positive infinity from both sides of a
f(x) decreases toward negative infinity from both sides of a
Worked Example 5.1
Find the one-sided infinite limits of f(x) = 1 divided by (x minus 2) near x = 2.
As x approaches 2 from the right (x slightly greater than 2):
x minus 2 is a small positive number. 1 divided by a small positive → +∞
As x approaches 2 from the left (x slightly less than 2):
x minus 2 is a small negative number. 1 divided by a small negative → -∞
Vertical asymptote at x = 2. The left and right limits are different infinities, so the two-sided limit does not exist.
Vertical Asymptote Connection
For a rational function in reduced form (common factors already cancelled), vertical asymptotes occur where the denominator equals zero. If the factor cancels, you have a removable discontinuity (a hole), not a vertical asymptote.
6. Limit Laws
The limit laws allow you to break complex limits into simpler pieces. They hold whenever the individual limits on the right-hand side exist as finite numbers.
| Law | Statement |
|---|---|
| Sum Law | the limit of f(x) plus g(x) equals the limit of f(x) plus the limit of g(x) |
| Difference Law | the limit of f(x) minus g(x) equals the limit of f(x) minus the limit of g(x) |
| Constant Multiple Law | the limit of c times f(x) equals c times the limit of f(x) |
| Product Law | the limit of f(x) times g(x) equals the limit of f(x) times the limit of g(x) |
| Quotient Law | the limit of f(x) divided by g(x) equals the limit of f(x) divided by the limit of g(x), provided the limit of g(x) is not zero |
| Power Law | the limit of f(x) raised to the nth power equals the limit of f(x) raised to the nth power |
| Root Law | the limit of the nth root of f(x) equals the nth root of the limit of f(x) |
Worked Example 6.1 — Applying Multiple Laws
Find the limit as x approaches 3 of the expression (2x squared minus x plus 4).
= lim(2x squared) - lim(x) + lim(4) [Sum/Difference Laws]
= 2 times lim(x squared) - lim(x) + 4 [Constant Multiple Law]
= 2 times [lim(x)] squared - lim(x) + 4 [Power Law]
= 2 times (3) squared - 3 + 4 [Direct Substitution for polynomials]
= 18 - 3 + 4 = 19
7. Direct Substitution
When a function is continuous at x equals a, you can evaluate the limit by simply plugging in a. This is the fastest and most common method.
Direct Substitution Property
If f is a polynomial or rational function, and a is in the domain of f, then:
Example 7.1 — Polynomial
Limit as x approaches -2 of (x cubed plus 5x minus 1) = (-2) cubed + 5(-2) - 1 = -8 - 10 - 1 = -19.
Example 7.2 — Rational Function
Limit as x approaches 4 of (x squared plus 1) divided by (x minus 1) = (16 + 1) divided by (4 - 1) = 17 divided by 3.
Example 7.3 — When It Fails
Limit as x approaches 1 of (x squared minus 1) divided by (x minus 1): direct substitution gives (1 - 1) divided by (1 - 1) = 0/0. This is an indeterminate form — direct substitution fails. You must use another technique.
Recognizing 0/0
When direct substitution produces 0/0, do not stop and say the limit does not exist. This indeterminate form means “more work needed” — the limit likely does exist but requires factoring, rationalizing, or another technique. See Section 8.
8. Indeterminate Forms — Handling 0/0
When direct substitution yields 0/0, you need to transform the expression into an equivalent form that allows direct substitution. The three main techniques are factoring and canceling, rationalizing with the conjugate, and algebraic manipulation.
Factor the numerator and/or denominator, then cancel the common factor that creates the 0/0. The resulting expression is identical to the original everywhere except at x = a, so it has the same limit.
Example 8.1
Limit as x approaches 1 of (x squared minus 1) divided by (x minus 1).
Direct substitution: (1 - 1)/(1 - 1) = 0/0 — indeterminate form
Factor numerator: (x - 1)(x + 1) / (x - 1)
Cancel (x - 1): = x + 1, for x ≠ 1
Now substitute: lim(x → 1) of (x + 1) = 1 + 1
Limit = 2
Example 8.2 — Trinomial
Limit as x approaches -3 of (x squared plus x minus 6) divided by (x plus 3).
Direct substitution: 0/0
Factor: (x + 3)(x - 2) / (x + 3)
Cancel: = x - 2, for x ≠ -3
Limit = -3 - 2 = -5
When the expression contains a square root and produces 0/0, multiply numerator and denominator by the conjugate of the radical expression. Use the identity (a - b)(a + b) = a squared minus b squared to eliminate the radical in the denominator (or numerator).
Example 8.3
Limit as x approaches 0 of (the square root of x plus 4, minus 2) all divided by x.
Direct substitution: (sqrt(4) - 2)/0 = 0/0
Conjugate of numerator: (sqrt(x + 4) + 2)
Multiply top and bottom by (sqrt(x + 4) + 2):
Numerator: (x + 4) - 4 = x
Denominator: x times (sqrt(x + 4) + 2)
Cancel x: 1 / (sqrt(x + 4) + 2)
Substitute x = 0: 1 / (sqrt(4) + 2) = 1/(2 + 2)
Limit = 1/4
Example 8.4
Limit as x approaches 9 of (x minus 9) divided by (square root of x minus 3).
Direct substitution: 0/0
Multiply by conjugate (sqrt(x) + 3)/(sqrt(x) + 3):
Numerator: (x - 9)(sqrt(x) + 3)
Denominator: (sqrt(x))² - 9 = x - 9
Cancel (x - 9): = sqrt(x) + 3
Substitute x = 9: sqrt(9) + 3 = 3 + 3
Limit = 6
Some 0/0 indeterminate forms require combining fractions, expanding, or other algebraic steps before factoring and canceling become possible.
Example 8.5 — Compound Fraction
Limit as x approaches 0 of the quantity (1 divided by x plus 2) minus (one half), all divided by x.
Direct substitution: (1/2 - 1/2)/0 = 0/0
Combine the fraction in numerator over common denominator 2(x+2):
Numerator becomes: [2 - (x + 2)] / [2(x + 2)] = [-x] / [2(x + 2)]
Divide by x: [-x] / [2(x + 2)] divided by x = -1 / [2(x + 2)]
Substitute x = 0: -1 / [2(2)]
Limit = -1/4
9. The Squeeze Theorem
Squeeze Theorem (Sandwich Theorem)
Suppose that g(x) is less than or equal to f(x) is less than or equal to h(x) for all x near a (except possibly at a itself). If the limit of g(x) as x approaches a equals L, and the limit of h(x) as x approaches a also equals L, then:
f(x) is squeezed between two functions that share the same limit, so f must share that limit too.
Worked Example 9.1 — Classic Application
Show that the limit as x approaches 0 of x times sin(1/x) equals 0.
Because -1 is less than or equal to sin(1/x) is less than or equal to 1 for all x not equal to 0, we can multiply through by |x| (which is non-negative):
-|x| ≤ x sin(1/x) ≤ |x|
Now apply limits: the limit of -|x| as x approaches 0 = 0, and the limit of |x| as x approaches 0 = 0.
By the Squeeze Theorem, the limit of x sin(1/x) as x approaches 0 = 0.
Worked Example 9.2 — Special Trig Limit
The Squeeze Theorem is used to prove the fundamental trigonometric limit: the limit as x approaches 0 of sin(x) divided by x equals 1. This requires geometric bounding of the sine function between two triangle areas.
Result: lim(x → 0) sin(x)/x = 1. This is essential for finding the derivatives of sine and cosine in calculus.
10. Continuity — The Definition
Intuitively, a function is continuous if you can draw its graph without lifting your pencil. Formally, continuity at a point requires three conditions to hold simultaneously.
Three Conditions for Continuity at x = a
- 1f(a) is defined. (No hole or gap — the function has a value at x = a.)
- 2The limit of f(x) as x approaches a exists. (Both one-sided limits exist and are equal.)
- 3The limit equals the function value: lim(x → a) f(x) = f(a). (The function actually reaches what it approaches.)
Worked Example 10.1 — Testing Continuity
Is f(x) = (x squared minus 4) divided by (x minus 2) continuous at x = 2?
Condition 1: f(2) = (4 - 4)/(2 - 2) = 0/0. Undefined. Condition 1 fails.
Not continuous at x = 2. The function has a removable discontinuity (hole) there.
Note: the two-sided limit does exist (it equals 4), so the discontinuity is removable. We could make f continuous by defining f(2) = 4.
Worked Example 10.2 — Continuity on an Interval
Polynomial functions are continuous everywhere. Rational functions are continuous everywhere except where the denominator equals zero. Sine and cosine are continuous everywhere. Exponential and logarithmic functions are continuous on their domains.
Example: f(x) = (x plus 1) divided by (x squared minus 9) is continuous everywhere except at x = 3 and x = -3, where the denominator is zero and vertical asymptotes occur.
11. Types of Discontinuity
When at least one of the three continuity conditions fails, the function is discontinuous at that point. The type of discontinuity depends on how and which condition fails.
The limit exists at x equals a, but either f(a) is undefined or f(a) does not equal the limit. Graphically this is a hole in the curve.
Example
f(x) = (x squared minus 4) divided by (x minus 2) at x = 2. The limit as x approaches 2 is 4, but f(2) is undefined. Remove the discontinuity by defining f(2) = 4.
The left-hand limit and right-hand limit both exist but are not equal. The function jumps from one value to another. The two-sided limit does not exist.
Example
A piecewise function defined as x plus 1 for x less than 2, and x squared minus 1 for x greater than or equal to 2. At x = 2: left limit = 3, right limit = 3, but the function definition creates a jump at any piecewise boundary where the pieces do not meet.
The function grows without bound near x equals a — a vertical asymptote. At least one one-sided limit is positive or negative infinity, so the two-sided limit does not exist.
Example
f(x) = 1 divided by (x minus 3) at x = 3. As x approaches 3 from the right, f(x) goes to positive infinity. As x approaches 3 from the left, f(x) goes to negative infinity.
12. Intermediate Value Theorem
Intermediate Value Theorem (IVT)
If f is continuous on the closed interval from a to b, and N is any number strictly between f(a) and f(b) — that is, f(a) is less than N is less than f(b) (or vice versa) — then there exists at least one number c in the open interval from a to b such that f(c) = N.
A continuous function cannot skip values. To get from f(a) to f(b), it must pass through every value in between.
Worked Example 12.1 — Proving a Root Exists
Show that f(x) = x cubed minus x minus 1 has a root between x = 1 and x = 2.
f(1) = 1 - 1 - 1 = -1. Negative.
f(2) = 8 - 2 - 1 = 5. Positive.
f is continuous (it is a polynomial) on [1, 2].
Since f(1) = -1 is less than 0 is less than 5 = f(2), by the IVT there
exists at least one c in (1, 2) with f(c) = 0. The function must cross zero between x = 1 and x = 2.
Worked Example 12.2 — Finding a Specific Value
Show that f(x) = x squared takes the value 7 somewhere on the interval from 2 to 3.
f(2) = 4 and f(3) = 9.
4 is less than 7 is less than 9, and f is continuous on [2, 3].
By IVT, there exists c in (2, 3) with f(c) = 7, namely c = sqrt(7).
IVT Limitation
The IVT guarantees existence but not uniqueness — there may be more than one c satisfying f(c) = N. It also requires continuity; for discontinuous functions, the theorem does not apply and the conclusion may fail.
13. Epsilon-Delta Definition — Conceptual Introduction
The intuitive description of limits — “f(x) approaches L as x approaches a” — needs a precise mathematical definition for rigorous proofs. The epsilon-delta definition makes “approaches” exact.
Formal Definition
We say that the limit of f(x) as x approaches a equals L if: for every number epsilon greater than 0 (no matter how small), there exists a number delta greater than 0 such that whenever 0 is less than |x minus a| is less than delta, it follows that |f(x) minus L| is less than epsilon.
if 0 < |x - a| < δ, then |f(x) - L| < ε
What the Definition Means
Epsilon represents an “error tolerance” in the output — how close we require f(x) to be to L. Any positive epsilon is a valid challenge, no matter how tiny.
Delta represents the corresponding “input restriction” — how close x must be to a to guarantee the output is within epsilon of L. You must be able to find such a delta for every epsilon.
The condition 0 is less than |x minus a| means x is not equal to a. The limit is about the approach, not the value at the point.
Example 13.1 — Verifying a Limit with Epsilon-Delta
Prove that the limit as x approaches 3 of (2x minus 1) equals 5.
We need: for every ε greater than 0, find δ such that if 0 is less than |x - 3| is less than δ, then |(2x - 1) - 5| is less than ε.
Simplify the output condition:
|(2x - 1) - 5| = |2x - 6| = 2|x - 3|
We need 2|x - 3| is less than ε, which means |x - 3| is less than ε/2.
Choose δ = ε/2. Then whenever |x - 3| is less than δ = ε/2:
|(2x - 1) - 5| = 2|x - 3| < 2(ε/2) = ε
The proof is complete. The limit equals 5.
14. Connection to Derivatives — Difference Quotient Preview
Everything covered in this guide is preparation for the central concept of calculus: the derivative. The derivative measures instantaneous rate of change, and it is defined entirely in terms of a limit.
The Difference Quotient
The average rate of change of f over the interval from x to x + h is:
This is the slope of the secant line between two points on the graph. As h approaches 0, the two points merge, and the secant line becomes the tangent line. The derivative is the limit of this quotient:
Worked Example 14.1 — Derivative of x squared
Use the difference quotient to find the derivative of f(x) = x squared.
f'(x) = lim(h → 0) [(x + h)² - x²] / h
= lim(h → 0) [x² + 2xh + h² - x²] / h
= lim(h → 0) [2xh + h²] / h
= lim(h → 0) h(2x + h) / h
= lim(h → 0) (2x + h) [cancel h, indeterminate form resolved]
= 2x
Direct substitution would give 0/0. Instead, algebra cancels the h in the denominator before taking the limit — the same factoring/canceling technique from Section 8, applied to the difference quotient.
Worked Example 14.2 — Slope of Tangent Line
Find the slope of the tangent line to f(x) = x squared at the point (3, 9).
From Example 14.1: f'(x) = 2x
At x = 3: slope = 2(3) = 6
Tangent line: y - 9 = 6(x - 3), or y = 6x - 9
Why Limits Are the Foundation of Calculus
- The derivative is a limit of the difference quotient.
- The definite integral is a limit of Riemann sums.
- L'Hopital's rule evaluates indeterminate limits using derivatives.
- Series convergence is defined using limits of partial sums.
- Every major theorem in calculus (Mean Value Theorem, Fundamental Theorem) requires understanding limits and continuity.
Frequently Asked Questions
What is the intuitive idea of a limit in precalculus?
A limit describes the value that a function approaches as x gets arbitrarily close to some number a, without necessarily reaching it. If you trace the graph of f(x) from both sides toward x equals a, the value your finger approaches is the limit. The key idea: the function does not need to be defined at a for the limit to exist.
What is the difference between a left-hand limit and a right-hand limit?
The left-hand limit (written as x approaches a from the left, using a superscript minus) is the value f(x) approaches as x increases toward a. The right-hand limit (x approaches a from the right, superscript plus) is the value f(x) approaches as x decreases toward a. For the two-sided limit to exist, both one-sided limits must exist and be equal.
What are indeterminate forms and how do you handle 0/0?
An indeterminate form like 0/0 arises when direct substitution gives a fraction with zero in both numerator and denominator. This does not mean the limit is undefined — it means you need another technique. The three main methods are: (1) factor and cancel the common factor, (2) rationalize using the conjugate for expressions with square roots, and (3) expand algebraically. After simplifying, apply direct substitution to the resulting expression.
What is the Squeeze Theorem?
The Squeeze Theorem states: if g(x) is less than or equal to f(x) is less than or equal to h(x) for all x near a, and if the limit of g(x) as x approaches a equals L and the limit of h(x) as x approaches a equals L, then the limit of f(x) as x approaches a also equals L. The classic application is proving that the limit of x times sin(1/x) as x approaches 0 equals 0, by squeezing it between -|x| and |x|.
What does it mean for a function to be continuous at a point?
A function f is continuous at x equals a if three conditions hold: (1) f(a) is defined, (2) the limit of f(x) as x approaches a exists, and (3) the limit equals f(a). If any one of these conditions fails, the function has a discontinuity at that point. Polynomial and rational functions (away from zeros of the denominator) are continuous everywhere on their domains.
What is the Intermediate Value Theorem?
The Intermediate Value Theorem (IVT) states: if f is continuous on the closed interval from a to b, and N is any value strictly between f(a) and f(b), then there exists at least one c in the open interval from a to b such that f(c) equals N. In plain language: a continuous function cannot skip values. If it starts at 3 and ends at 7, it must hit every value between 3 and 7 at some point.
How does the limit connect to the derivative?
The derivative is defined as the limit of the difference quotient: the limit as h approaches 0 of the quantity f(x plus h) minus f(x), all divided by h. This is the slope of the secant line between two points becoming a tangent line slope as the two points merge together. Understanding limits is the essential prerequisite for understanding what a derivative actually means.
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