Difficulty: Introductory to Intermediate | Prerequisites: Overview notes (functions at four levels), basic algebra (factoring, rationalising).
These three sections build the precise machinery you need before derivatives can be defined rigorously. §1.4 poses the two motivating problems (velocity and tangent slope), §1.5 introduces the limit as the tool that solves them, and §1.6 gives you the algebraic rules (Limit Laws) that make limit computations practical. If you are coming into this cold: a limit captures the value a function "wants to approach" near a point, even if it never reaches it. Every definition in the rest of the course rests on this idea.
Instantaneous velocity and tangent slope are both defined as limits of average rates over shrinking intervals. A limit lim(x→a) f(x) = L means f(x) can be forced as close to L as you like by keeping x close enough to a. The Limit Laws let you break complicated limits into simple pieces, and the Squeeze Theorem handles wildly oscillating functions.
Average velocity
The total distance travelled divided by the total time elapsed over an interval: v_avg = [f(b) − f(a)] / (b − a). Think of it as the overall pace between two moments.
Instantaneous velocity
The velocity at a single instant, defined as the limit of average velocities over shorter and shorter intervals: v = lim(h→0) [f(a+h) − f(a)] / h. This is the speedometer reading at one moment.
Secant line
A straight line cutting through a curve at two points (a, f(a)) and (b, f(b)). Its slope equals the average rate of change over [a, b].
Tangent line
The line that just touches the curve at one point (a, f(a)), defined as the limit of secant lines as the second point slides toward the first. Its slope is the instantaneous rate of change.
Limit
lim(x→a) f(x) = L means f(x) can be forced arbitrarily close to L by making x sufficiently close to (but not equal to) a. In plain terms, L is the value f(x) "wants to be" near a, regardless of what f(a) itself is.
Continuous (at a point)
A function f(x) is continuous at x = a when lim(x→a) f(x) = f(a). The graph passes through (a, f(a)) with no jump, hole or break.
One-sided limit
A limit that considers x approaching a from one direction only. lim(x→a⁺) f(x) = L (from the right) or lim(x→a⁻) f(x) = L (from the left). The two-sided limit exists only when both one-sided limits agree.
Infinite limit
lim(x→a) f(x) = ∞ means f(x) grows without bound as x approaches a. The graph has a vertical asymptote at x = a.
Squeeze Theorem
If f(x) ≤ g(x) ≤ h(x) near a, and lim(x→a) f(x) = lim(x→a) h(x) = L, then lim(x→a) g(x) = L. In simple terms, if a function is trapped between two others that both approach L, it must approach L too.
The velocity problem
A particle at position f(t) at time t has average velocity v_avg = [f(b) − f(a)] / (b − a) over the interval [a, b]. The instantaneous velocity at t = a is defined as the limit:
v = lim(h→0) [f(a+h) − f(a)] / h
Worked example: falling stone
f(t) = 16t². Average velocity from t = 3 to t = 4:
v_avg = [f(4) − f(3)] / 1 = [256 − 144] / 1 = 112 ft/sec
Average velocity from t = 3 to t = 3.1:
v_avg = [16(3.1²) − 16(3²)] / 0.1 = [153.76 − 144] / 0.1 = 97.6 ft/sec
As h shrinks (1, 0.1, 0.01, 0.001, ...), the averages approach 96. The algebraic proof:
v_avg = [16(3+h)² − 16(9)] / h = 16[(9 + 6h + h²) − 9] / h = 16(6h + h²) / h = 96 + 16h
As h → 0, the error term 16h vanishes, leaving v = 96 ft/sec exactly.
The tangent problem
The slope of the secant line through (a, f(a)) and (b, f(b)) is m_sec = [f(b) − f(a)] / (b − a), which equals the average velocity. As the second point slides toward the first (b → a, or equivalently h → 0), the secant line approaches the tangent line. The tangent slope is:
m = lim(h→0) [f(a+h) − f(a)] / h
This is exactly the same formula as instantaneous velocity. For f(t) = 16t² at t = 3, the tangent line is:
y = 144 + 96(x − 3)
The definition
lim(x→a) f(x) = L means f(x) can be forced arbitrarily close to L by making x close enough to a (but x ≠ a). The value f(a) itself is irrelevant; the limit only cares about what happens near a.
Evaluating limits: three scenarios
Continuous case (just plug in). If substituting x = a gives a well-defined value, that value is the limit. Example: lim(x→5) x² = 25.
Limit exists but function is undefined or "wrong" at a. Example: lim(x→1) (x² − 1)/(x − 1). Plugging in gives 0/0, but factoring: (x−1)(x+1)/(x−1) = x + 1, so the limit is 2. The function has a hole at x = 1, yet the limit still exists.
Limit does not exist (DNE). The function does not settle on a single value. Example: lim(x→0) sgn(x) does not exist because sgn(x) = +1 for x > 0 and −1 for x < 0, with no single value to approach.
One-sided limits
lim(x→a⁺) f(x) considers only x > a (approaching from the right). lim(x→a⁻) f(x) considers only x < a (from the left). The two-sided limit exists if and only if both one-sided limits exist and are equal.
For sgn(x): lim(x→0⁺) sgn(x) = 1 and lim(x→0⁻) sgn(x) = −1. Since these disagree, the two-sided limit does not exist.
Infinite limits
lim(x→a) f(x) = ∞ means f(x) grows without bound near a. The line x = a is a vertical asymptote.
lim(x→0) 1/|x| = ∞ (both sides blow up to +∞).
lim(x→0⁺) 1/x = ∞ and lim(x→0⁻) 1/x = −∞ (opposite sides go opposite ways, so the two-sided limit is neither +∞ nor −∞).
The Limit Laws
If lim(x→a) f(x) and lim(x→a) g(x) both exist, then:
Sum: lim(x→a) [f(x) + g(x)] = lim f(x) + lim g(x)
Difference: lim(x→a) [f(x) − g(x)] = lim f(x) − lim g(x)
Constant Multiple: lim(x→a) [c · f(x)] = c · lim f(x)
Product: lim(x→a) [f(x) · g(x)] = lim f(x) · lim g(x)
Quotient: lim(x→a) [f(x)/g(x)] = lim f(x) / lim g(x), provided lim g(x) ≠ 0
Power: lim(x→a) [f(x)]ⁿ = [lim f(x)]ⁿ
Root: lim(x→a) ⁿ√f(x) = ⁿ√[lim f(x)] (if n is even, require lim f(x) > 0)
The pattern: the limit of an operation equals the operation applied to the limits. All laws hold for one-sided limits too.
Limits by plugging in
Combining the laws with the basic limits lim(x→a) x = a and lim(x→a) c = c shows that most functions built from polynomials, roots and trig are continuous. So for these, lim(x→a) f(x) = f(a), i.e., just substitute.
Limits by cancelling zeroes (the 0/0 case)
When substitution gives 0/0, factor out and cancel the vanishing term before substituting.
Example: lim(x→2) (x² − 4x + 4)/(x² − x − 2) = lim(x→2) (x−2)²/[(x−2)(x+1)] = lim(x→2) (x−2)/(x+1) = 0/3 = 0.
Conjugate radical trick: lim(x→9) (x−9)/(√x − 3). Multiply top and bottom by (√x + 3): the numerator becomes (x−9)(√x+3), the denominator becomes x − 9, and after cancellation the limit is √9 + 3 = 6.
Limits by cases
When a function is defined piecewise, compute the left and right limits separately. If they agree, the two-sided limit exists.
Example: lim(x→0) √(1 + sgn(x)²). From the right, sgn(x) = 1, so the limit is √2. From the left, sgn(x) = −1, giving √(1+1) = √2 again. Both sides agree, so the limit is √2.
The Squeeze Theorem
If f(x) ≤ g(x) ≤ h(x) near a, and lim(x→a) f(x) = lim(x→a) h(x) = L, then lim(x→a) g(x) = L.
Classic application: lim(x→0) x sin(π/x). The function sin(π/x) oscillates wildly, so the Product Law fails. But −|x| ≤ x sin(π/x) ≤ |x|, and both bounds go to 0, so by the Squeeze Theorem the limit is 0.
Note: lim(x→0) sin(π/x) on its own does not exist, because the oscillations never settle.
Average velocity / secant slope
v_avg = m_sec = [f(b) − f(a)] / (b − a)
Instantaneous velocity / tangent slope
v = m = lim(h→0) [f(a+h) − f(a)] / h
Tangent line equation (point-slope form)
y = f(a) + m(x − a), where m = lim(h→0) [f(a+h) − f(a)] / h
Limit Laws summary
Sum, Difference, Constant Multiple, Product, Quotient, Power, Root: "the limit of an operation = the operation of the limits."
Squeeze Theorem
f(x) ≤ g(x) ≤ h(x) near a, and lim f(x) = lim h(x) = L ⟹ lim g(x) = L
Every time your car's speedometer reads 60 mph, it is reporting an instantaneous velocity, the derivative of your position with respect to time. GPS navigation systems compute average velocities over short intervals and take limits computationally to estimate your current speed. The tangent-line idea is also used in engineering to linearise nonlinear systems: near an operating point, a complicated curve behaves like its tangent line, which is far simpler to analyse.
Students confuse "the limit equals L" with "f(a) equals L." The limit depends only on values near a, never on f(a) itself. A function can have a limit at a point where it is undefined or has a different value.
A common error is to assume a limit does not exist whenever substitution gives 0/0. The expression 0/0 is indeterminate, meaning the limit might exist (and often does) after algebraic simplification.
Students sometimes apply the Quotient Law even when the denominator's limit is zero. The law requires lim g(x) ≠ 0; when it is zero, you need cancellation or another technique.
The Squeeze Theorem is not a last resort. It is often the only clean way to evaluate limits involving oscillating functions. Recognise the pattern: a bounded oscillation multiplied by something going to zero.
⚠️ The definition of a limit (§1.5) is the conceptual foundation of the entire course. You will be asked to state it and explain what it means.
⚠️ Factoring to resolve 0/0 indeterminate forms is one of the most frequently tested skills. Practise recognising common factorisations (difference of squares, perfect square trinomials).
⚠️ One-sided limits and their relationship to two-sided limits appear regularly. Know: the two-sided limit exists if and only if both one-sided limits exist and agree.
⚠️ The Squeeze Theorem is a classic exam question. The typical setup is x · sin(1/x) or x² · cos(1/x) near x = 0.
True or False: lim(x→a) f(x) can exist even when f(a) is undefined.
True. The limit only depends on values near a, not at a.
Fill in the blank: The slope of the secant line through (a, f(a)) and (b, f(b)) equals the ______ velocity over [a, b].
Average.
True or False: If lim(x→0⁺) f(x) = 3 and lim(x→0⁻) f(x) = 5, then lim(x→0) f(x) = 4.
False. The two-sided limit does not exist because the one-sided limits disagree.
What technique do you use when direct substitution in a limit gives 0/0?
Factor and cancel (or multiply by a conjugate radical), then substitute again.
Q: Compute lim(x→3) (x² − 9)/(x − 3).
A: Factor the numerator: (x−3)(x+3)/(x−3) = x + 3. Then lim(x→3) (x+3) = 6.
Q: Does lim(x→0) (1/x) exist? Explain.
A: No. lim(x→0⁺) 1/x = +∞ and lim(x→0⁻) 1/x = −∞. The one-sided limits do not agree (and are not finite), so the limit does not exist.
Q: Use the Squeeze Theorem to evaluate lim(x→0) x² cos(1/x).
A: Since −1 ≤ cos(1/x) ≤ 1, we have −x² ≤ x² cos(1/x) ≤ x². Both −x² and x² approach 0 as x → 0. By the Squeeze Theorem, the limit is 0.
Q: Find the tangent line to y = x³ at the point (2, 8).
A: m = lim(h→0) [(2+h)³ − 8] / h = lim(h→0) [12h + 6h² + h³] / h = lim(h→0) (12 + 6h + h²) = 12. Tangent line: y = 8 + 12(x − 2), i.e. y = 12x − 16.
Q: Evaluate lim(x→4) (√x − 2)/(x − 4) using the conjugate radical method.
A: Multiply by (√x + 2)/(√x + 2): the numerator becomes x − 4, and the expression simplifies to 1/(√x + 2). As x → 4, this is 1/(2 + 2) = 1/4.
The limit concept introduced here is formalised with the epsilon-delta definition in §1.7. Continuity (§1.8) is defined directly in terms of limits. Every derivative formula in Chapters 2 and 3 rests on evaluating a limit of a difference quotient. The Squeeze Theorem reappears when proving lim(x→0) (sin x)/x = 1, a result needed for differentiating trig functions.
limits, limit laws, squeeze theorem, sandwich theorem, tangent line, secant line, instantaneous velocity, average velocity, difference quotient, one-sided limits, infinite limits, vertical asymptote, continuous, discontinuity, 0/0 indeterminate form, conjugate radical, factoring limits, Stewart calculus §1.4 §1.5 §1.6, MATH 132, MSU calculus