Difficulty: Introductory. Prerequisites: college algebra, basic graphing of functions.
Limits describe what value a function approaches as the input gets close to some number, even if the function is not defined there. One-sided limits look at the approach from just one direction (left or right), and the two-sided limit exists only when both sides agree. Continuity means the function has no jumps, holes or breaks: the limit equals the function's value at that point.
Limit
The value that f(x) approaches as x gets closer and closer to some number c. Written as lim(x→c) f(x). In simple terms, it is the y-value the graph is heading toward, regardless of whether it ever gets there.
One-sided limit (left-hand limit)
lim(x→c⁻) f(x) = K. The value f(x) approaches as x comes from values less than c (from the left on the number line). Think of it as: what happens to the y-values if you walk along the graph toward x = c from the left side only.
One-sided limit (right-hand limit)
lim(x→c⁺) f(x) = L. The value f(x) approaches as x comes from values greater than c (from the right). Same idea, opposite direction.
Two-sided limit
lim(x→c) f(x) = L. Exists if and only if the left-hand and right-hand limits are both equal to L. In simple terms, the function has to be heading to the same y-value from both sides for the two-sided limit to exist.
Indeterminate form (0/0)
When direct substitution into a limit gives 0/0, the limit might still exist. You need algebraic work (factoring, rationalising) to find out. Think of it as: the answer is hidden, not missing.
Non-zero over zero form (n/0)
When substitution gives a non-zero number divided by zero, the limit does not exist as a finite number. It will be +∞, −∞, or DNE (if the two sides disagree on sign).
Vertical asymptote
A vertical line x = a where the function blows up to ±∞. Found by identifying zeros of the denominator that do not cancel. In simple terms, it is a wall the graph climbs or falls along but never crosses.
Horizontal asymptote
A horizontal line y = L that the graph approaches as x → +∞ or x → −∞. Found by evaluating the limit at infinity. Think of it as the long-run levelling-off value of the function.
Continuity at a point
A function f(x) is continuous at x = c when three conditions hold: (I) f(c) is defined, (II) lim(x→c) f(x) exists, (III) lim(x→c) f(x) = f(c). In simple terms, you can draw the graph through that point without lifting your pencil.
Hole (removable discontinuity)
A point where the limit exists but the function is either undefined or defined at a different value. The graph has a gap you could fill in with a single point.
To evaluate a one-sided limit from a graph, trace along the curve toward x = c from the specified side and read off the y-value the curve is heading toward.
The left-hand limit (x → c⁻) uses the part of the graph where x < c.
The right-hand limit (x → c⁺) uses the part where x > c.
f(c) is the actual y-value at x = c (the filled dot, if one exists). A limit can exist even when f(c) does not, and f(c) can exist even when the limit does not.
If the left-hand and right-hand limits are different, the two-sided limit Does Not Exist (DNE).
Direct substitution: plug x = c into f(x). If you get a real number, that is the limit.
Works immediately for polynomials (e.g., lim(x→3) (x + 5) = 8) and for rational functions where the denominator is not zero at c.
One-sided limits are evaluated algebraically the same way; the direction only matters when the function has a piecewise definition or when you need to determine the sign near c.
If direct substitution gives 0/0, the expression is indeterminate. Factor, cancel, or rationalise to simplify, then substitute again.
Example: lim(x→1) (x² − 1)/(x − 1). Direct substitution gives 0/0. Factor the numerator: (x − 1)(x + 1)/(x − 1) = x + 1. Then lim(x→1)(x + 1) = 2.
For square-root expressions, multiply by the conjugate. Example: lim(x→64) (√x − 8)/(x − 64). Multiply top and bottom by (√x + 8) to clear the radical.
If substitution gives n/0 where n ≠ 0, the limit is infinite. You must check the sign from each side using a table of values.
If both sides go to +∞, the limit is +∞. If both go to −∞, the limit is −∞. If they disagree, the two-sided limit is DNE.
Example: lim(x→1) 1/(x − 1). From the left the values go to −∞; from the right, to +∞. The two-sided limit is DNE.
Polynomial functions: lim(x→±∞) of a polynomial is determined entirely by the leading term. Ignore every other term.
Example: lim(x→−∞)(−2x⁵ − 4x³ + 2) = lim(x→−∞)(−2x⁵) = +∞ (odd power, negative coefficient, x negative).
Rational functions: compare the degree of the numerator (m) and the denominator (n).
m < n: the limit is 0.
m = n: the limit is the ratio of the leading coefficients, a_m / b_n.
m > n: the limit is ±∞ (no horizontal asymptote).
Vertical asymptote at x = a: set the denominator equal to zero and solve. Check that the 0/0 form does not appear (if it does, the factor cancels and you may have a hole instead).
Horizontal asymptote y = L: evaluate lim(x→∞) f(x) using the leading-term method.
Holes: if both numerator and denominator share a common factor (x − a), the function simplifies and x = a is a hole, not a vertical asymptote. The coordinates of the hole are (a, simplified function evaluated at a).
Example: f(x) = (x² − 2x − 3)/(x² − 1) = (x − 3)(x + 1)/((x − 1)(x + 1)). The (x + 1) cancels. Hole at x = −1, vertical asymptote at x = 1, horizontal asymptote y = 1.
A function is continuous at x = c when all three conditions hold:
(I) f(c) is defined.
(II) lim(x→c) f(x) exists.
(III) lim(x→c) f(x) = f(c).
A function is continuous on an open interval (a, b) if it is continuous at every point in that interval.
For piecewise functions, check continuity at the boundary value by evaluating the limit from each side and comparing with the function value.
Example: f(x) = {x² − 3 if x < 1, Ax + 5 if x ≥ 1}. For continuity at x = 1, the left-hand limit (1² − 3 = −2) must equal f(1) = A + 5. So A = −7.
\text{Two-sided limit exists} \iff \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = L\text{Indeterminate form: } \lim_{x \to c} \frac{f(x)}{g(x)} = \frac{0}{0} \implies \text{simplify, then re-evaluate}\text{Rational function at infinity: } \lim_{x \to \pm\infty} \frac{a_m x^m + \cdots}{b_n x^n + \cdots} = \lim_{x \to \pm\infty} \frac{a_m x^m}{b_n x^n}Leading-term shortcut for rational functions:
Degree comparison | Limit | Horizontal asymptote |
|---|---|---|
m < n (numerator degree smaller) | 0 | y = 0 |
m = n (degrees equal) | a_m / b_n | y = a_m / b_n |
m > n (numerator degree larger) | ±∞ | None |
\text{Reciprocal limit: } \lim_{x \to \pm\infty} \frac{k}{x^p} = 0 \quad (p > 0,\; k \neq 0)\text{Continuity test at } x = c: \quad (I)\; f(c) \text{ defined}, \quad (II)\; \lim_{x \to c} f(x) \text{ exists}, \quad (III)\; \lim_{x \to c} f(x) = f(c)Students often think the limit of f(x) at x = c is the same as f(c). It is not. The limit describes where the function is heading; f(c) is the actual value at that point. These can differ (and the limit can exist when f(c) does not).
Students often assume 0/0 means "the limit does not exist." It does not mean that. 0/0 is indeterminate, which means more work is needed: the limit may exist, may not, or may be infinite.
Students sometimes forget to check both sides when the substitution gives n/0. You cannot conclude +∞ or −∞ without examining left and right behaviour separately.
Students confuse a hole with a vertical asymptote. A hole occurs when a common factor cancels from numerator and denominator. A vertical asymptote occurs when the denominator is zero but no cancellation is possible.
⚠️ Reading one-sided limits from a graph is a near-certain exam question. Practise quickly identifying whether a filled or open dot matters.
⚠️ Factoring to resolve 0/0 indeterminate forms is heavily tested. Be comfortable factoring quadratics and using conjugate multiplication for radicals.
⚠️ Identifying vertical and horizontal asymptotes from the function (not a graph) appears on most exams. Know the leading-term shortcut cold.
⚠️ The three-part continuity test is a common "show that" question. You will be asked to verify each condition, or to find a constant (like A) that makes a piecewise function continuous.
⚠️ Distinguishing holes from vertical asymptotes (does the factor cancel?) is a classic exam trap.
True or False: If lim(x→2⁻) f(x) = 5 and lim(x→2⁺) f(x) = 3, then lim(x→2) f(x) = 5. False. The two-sided limit does not exist because the one-sided limits are not equal.
Fill in the blank: When direct substitution gives 0/0, the form is called ________. Indeterminate.
True or False: A function can have a limit at x = c even if f(c) is undefined. True. The limit depends on nearby behaviour, not the value at c itself.
Fill in the blank: For the rational function f(x) = (3x² + 1)/(5x² − 2), the horizontal asymptote is y = ________. 3/5 (ratio of leading coefficients).
True or False: If the denominator of a rational function is zero at x = a, then x = a is always a vertical asymptote. False. If the numerator is also zero at x = a and the factor cancels, then x = a is a hole.
Q: Evaluate lim(x→6) (x² − 4x − 12)/(x − 6).
A: Factor the numerator: (x − 6)(x + 2)/(x − 6) = x + 2. Substituting x = 6 gives 8.
Q: Evaluate lim(x→−∞) (−2x⁴ − 3x + 1) / (x² − 5)².
A: The denominator expands to degree 4. Leading terms: −2x⁴ / x⁴ = −2.
Q: Given f(x) = {x² + x if x < 2, 7 if x = 2, 3 − x if x > 2}, does lim(x→2) f(x) exist?
A: Left-hand limit: (2)² + 2 = 6. Right-hand limit: 3 − 2 = 1. The one-sided limits differ, so the two-sided limit does not exist (DNE).
Q: Find all vertical and horizontal asymptotes of f(x) = (x² − 2x − 3)/(x² − 1). Are there any holes?
A: Factor: (x − 3)(x + 1) / ((x − 1)(x + 1)). The (x + 1) cancels, so there is a hole at x = −1. The simplified function is (x − 3)/(x − 1). Vertical asymptote at x = 1. Horizontal asymptote: degrees equal, so y = 1. The hole's y-coordinate is (−1 − 3)/(−1 − 1) = −4/−2 = 2, giving hole at (−1, 2).
Q: Show that f(x) = (x − 2)/(x + 1) is continuous at x = 1.
A: (I) f(1) = (1 − 2)/(1 + 1) = −1/2, which is defined. (II) lim(x→1) (x − 2)/(x + 1) = −1/2, which exists. (III) The limit equals f(1). All three conditions hold, so f is continuous at x = 1.
Limits are the foundation for the derivative (Sections 2.4–2.5), which is defined as a limit of a difference quotient. Every differentiation rule traces back to this idea. Continuity matters again when applying the Extreme Value Theorem (Section 4.5), which guarantees that a continuous function on a closed interval has an absolute max and min. Vertical and horizontal asymptotes reappear in the graphing strategy (Section 4.4), where you sketch rational functions using first and second derivatives.
Limits, one-sided limits, left-hand limit, right-hand limit, two-sided limit, limit existence, limit does not exist, DNE, indeterminate form, 0/0, n/0, infinite limits, vertical asymptote, horizontal asymptote, hole, removable discontinuity, continuity, continuous function, three-part continuity test, piecewise function, leading-term method, limits at infinity, reciprocal limits, MAC 2233, business calculus, FIU, calculus for business