Difficulty: Intermediate | Prerequisites: Limits and limit laws (Section 2.1), basic algebra (factoring, rationalisation, common denominators)
The limit definition of the derivative is the formula f'(x) = lim(h→0) [f(x+h) - f(x)] / h. You simplify the difference quotient algebraically until you can cancel the h in the denominator, then take the limit as h→0. Once you have f'(x), you can evaluate it at a specific point and write the equation of the tangent line there.
Difference quotient
The expression [f(x+h) - f(x)] / h. It represents the average rate of change of f between x and x+h, which is the slope of the secant line through those two points.
Think of it as: the rise-over-run between two points on the curve, where the second point is h units to the right of the first.
Limit definition of the derivative
f'(x) = lim(h→0) [f(x+h) - f(x)] / h. You take the difference quotient and shrink h to zero, turning the secant slope into the tangent slope.
In simple terms, this is the precise way to find the instantaneous slope at any point on a curve.
Derivative (as a function)
The result of the limit definition, f'(x), is itself a function of x. It tells you the slope of f at every x-value where the limit exists. Also written df/dx.
Think of it as: a new function that answers "what is the slope of f right here?" for any x you plug in.
Tangent line
The line that just touches the curve at a single point and has the same slope as the curve at that point. Its equation is y = f(a) + f'(a)(x - a), where a is the point of tangency.
In simple terms, the best straight-line approximation to the curve at that spot.
Instantaneous rate of change
Another name for the derivative at a point. If s(t) is position, then s'(t) is velocity: the instantaneous rate of change of position with respect to time.
Think of it as: the speedometer reading at one exact moment, as opposed to the average speed over a trip.
Every problem in this assignment (Problems 3 to 9) follows the same four steps:
Write f(x+h), expand or simplify, then subtract f(x) and divide by h. This is part (a): the simplified difference quotient.
Take the limit as h→0. This is part (b): the derivative f'(x) or df/dx.
Evaluate the derivative at a specific x-value. This is part (c).
Use the point and the slope to write the tangent line equation y = f(a) + f'(a)(x - a). This is part (d).
The algebra in step 1 is where the work lives. The technique depends on the type of function.
f(x+h) = 7(x+h) - 6 = 7x + 7h - 6
Difference quotient: [7x + 7h - 6 - (7x - 6)] / h = 7h / h = 7
The h cancels immediately. The derivative is f'(x) = lim(h→0) 7 = 7. The derivative of any linear function mx + b is simply m. This makes sense: a line has constant slope everywhere.
At x = -3: f'(-3) = 7. And f(-3) = 7(-3) - 6 = -27. Tangent line: y = -27 + 7(x - (-3)) = 7x - 6. The tangent line to a line is the line itself.
f(x+h) = (x+h)² + 2(x+h) + 4 = x² + 2xh + h² + 2x + 2h + 4
Difference quotient: [x² + 2xh + h² + 2x + 2h + 4 - (x² + 2x + 4)] / h = [2xh + h² + 2h] / h = 2x + h + 2
The key step is expanding (x+h)², then everything without an h cancels, and every surviving term has an h you can factor out.
Derivative: f'(x) = lim(h→0) (2x + h + 2) = 2x + 2.
At x = 0: df/dx = 2(0) + 2 = 2. And f(0) = 4. Tangent line: y = 4 + 2(x - 0) = 2x + 4.
Problem 5: f(x) = √(x - 4). Problem 7: f(x) = √(x) + 4. Both use the same technique: rationalisation.
For f(x) = √(x - 4):
Difference quotient: [√(x+h - 4) - √(x - 4)] / h
Multiply numerator and denominator by the conjugate [√(x+h - 4) + √(x - 4)]:
= [(x+h - 4) - (x - 4)] / {h [√(x+h - 4) + √(x - 4)]} = h / {h [√(x+h - 4) + √(x - 4)]} = 1 / [√(x+h - 4) + √(x - 4)]
Derivative: f'(x) = lim(h→0) 1 / [√(x - 4) + √(x - 4)] = 1 / [2√(x - 4)].
At x = 13: f'(13) = 1 / [2√9] = 1/6. And f(13) = √9 = 3. Tangent line: y = 3 + (1/6)(x - 13) = (1/6)x + 5/6.
For f(x) = √x + 4, the same conjugate method gives f'(x) = 1 / (2√x). At x = 25: f'(25) = 1/10. And f(25) = 9. Tangent line: y = 9 + (1/10)(x - 25) = (1/10)x + 13/2.
These require finding a common denominator in the numerator, then simplifying.
Problem 6: f(t) = 2 / (t + 6)
Difference quotient: [2/(t+h+6) - 2/(t+6)] / h
Combine the fractions in the numerator using a common denominator of (t+h+6)(t+6):
= [2(t+6) - 2(t+h+6)] / [h(t+h+6)(t+6)] = [-2h] / [h(t+h+6)(t+6)] = -2 / [(t+h+6)(t+6)]
Derivative: f'(t) = lim(h→0) -2 / [(t+6)(t+6)] = -2 / (t+6)².
At t = -1: f'(-1) = -2/25. And f(-1) = 2/5. Tangent line: y = 2/5 + (-2/25)(t - (-1)) = (-2/25)t + 8/25.
Problem 8: r(t) = 6/t - t
Handle each piece: r(t+h) = 6/(t+h) - (t+h).
Difference quotient: [6/(t+h) - (t+h) - 6/t + t] / h = [6/(t+h) - 6/t - h] / h
Combine the rational part: [6t - 6(t+h)] / [ht(t+h)] - 1 = [-6h] / [ht(t+h)] - 1 = -6/[t(t+h)] - 1.
Derivative: r'(t) = lim(h→0) [-6/t² - 1] = -6/t² - 1.
At t = 1: r'(1) = -6 - 1 = -7. And r(1) = 6 - 1 = 5. Tangent line: y = 5 + (-7)(t - 1) = -7t + 12.
Problem 9: s(t) = (t+2)/(8-t) (a position function).
Difference quotient: [(t+h+2)/(8-t-h) - (t+2)/(8-t)] / h
Common denominator (8-t-h)(8-t), then expand numerator:
(t+h+2)(8-t) - (t+2)(8-t-h) = 8t - t² + 8h - th + 16 - 2t - [8t - t² - th + 16 - 2t - h(t+2)] = 8h + h(t+2) = h(10) ... let me redo: = (t+h+2)(8-t) - (t+2)(8-t-h).
Expand first: 8t + 8h - t² - th + 16 + 2(8) - 2t ... More carefully:
(t+h+2)(8-t) = 8t + 8h - t² - th + 16 + 2·8 - 2t ... Wait, let me be precise.
(t+h+2)(8-t) = (t+2)(8-t) + h(8-t)
(t+2)(8-t-h) = (t+2)(8-t) - h(t+2)
Subtract: h(8-t) + h(t+2) = h(8-t+t+2) = 10h
So the difference quotient simplifies to 10h / [h(8-t-h)(8-t)] = 10 / [(8-t-h)(8-t)].
Derivative: s'(t) = 10 / (8-t)².
Velocity at t = 7: v(7) = 10 / (8-7)² = 10 m/s.
Limit definition of the derivative
f'(x) = lim(h→0) [f(x+h) - f(x)] / h
Alternate notation: df/dx = lim(h→0) [f(x+h) - f(x)] / h
Tangent line equation at x = a
y = f(a) + f'(a)(x - a)
This is just point-slope form: you need the point (a, f(a)) and the slope f'(a).
Rationalisation identity (for square roots)
(√A - √B)(√A + √B) = A - B
Multiply the difference quotient's numerator and denominator by the conjugate to eliminate the square roots. After cancelling h, the limit becomes straightforward.
Common denominator technique (for rational functions)
When f involves a fraction, write the difference quotient with a single combined denominator. The h in the numerator will factor out and cancel with the h in the overall denominator, leaving an expression where you can set h = 0.
Problem 9 in the assignment frames this directly: s(t) = (t+2)/(8-t) is the position of an object at time t, and the derivative s'(t) is the velocity. This is the core idea behind kinematics in physics. Every time you see a speedometer, you are reading an instantaneous rate of change.
More broadly, derivatives model any quantity that changes: the rate at which a chemical reaction proceeds, the marginal cost of producing one more unit in economics, the rate of temperature change in engineering. The limit definition is the foundation for all of it.
Students often forget to subtract the entire f(x) when forming the difference quotient. With f(x) = x² + 2x + 4, the subtraction is -(x² + 2x + 4), not -x² + 2x + 4. Missing the parentheses is the single most common algebraic error.
Students sometimes try to "plug in h = 0" before simplifying. That gives 0/0, which is undefined. The whole point of the algebra is to cancel the h first, so the limit can then be evaluated.
With square root functions, students try to expand or factor the numerator directly. That does not work. The technique is rationalisation: multiply by the conjugate. If you see a square root difference, reach for the conjugate.
With rational functions, students sometimes invert incorrectly or lose signs when combining fractions. Write out the common denominator step carefully and double-check the sign on every term.
Students confuse the derivative function f'(x) with the derivative at a point f'(a). The first is a formula in x; the second is a number you get by substituting a specific value.
⚠️ The limit definition of the derivative is tested directly on MTH 132 exams. You will be given a function and asked to find the derivative from the definition, not using shortcut rules. Show every algebra step.
⚠️ "Find the equation of the tangent line" is a near-guaranteed exam question. Know the formula y = f(a) + f'(a)(x - a) cold.
⚠️ Square root and rational function derivatives from the definition are the most commonly tested because they require non-trivial algebra (rationalisation or common denominators). Polynomial derivatives are considered warm-up.
⚠️ Problem 9 style (position → velocity via derivative) combines the limit definition with a physical interpretation. Expect a problem like this on the exam, often with units.
True or false: The difference quotient [f(x+h) - f(x)] / h gives the slope of the tangent line at x.
Fill in the blank: To find the derivative of f(x) = √(x+3) using the definition, you multiply by the ______ to rationalise.
True or false: The derivative of f(x) = 5x + 1 using the limit definition is 5x.
Fill in the blank: The tangent line formula is y = f(a) + ______ (x - a).
True or false: If s(t) gives position in metres and t is in seconds, then s'(t) is measured in m/s.
Answers: 1. False (it gives the slope of the secant line; the tangent slope requires the limit as h→0). 2. Conjugate. 3. False (it is 5, not 5x). 4. f'(a). 5. True.
Q: Using the limit definition, find the derivative of f(x) = 3x² - x + 2.
A: f(x+h) = 3(x+h)² - (x+h) + 2 = 3x² + 6xh + 3h² - x - h + 2. Difference quotient: (6xh + 3h² - h)/h = 6x + 3h - 1. Limit as h→0: f'(x) = 6x - 1.
Q: Using the limit definition, find the derivative of f(x) = 1/(x+1).
A: Difference quotient: [1/(x+h+1) - 1/(x+1)] / h = [(x+1) - (x+h+1)] / [h(x+h+1)(x+1)] = -h / [h(x+h+1)(x+1)] = -1/[(x+h+1)(x+1)]. Limit as h→0: f'(x) = -1/(x+1)².
Q: Find the equation of the tangent line to f(x) = √(x+3) at x = 1.
A: f'(x) = 1/[2√(x+3)] (by rationalisation). f'(1) = 1/[2√4] = 1/4. f(1) = √4 = 2. Tangent line: y = 2 + (1/4)(x - 1) = (1/4)x + 7/4.
Q: The position of an object is given by s(t) = 4t/(t+1). Find the velocity at t = 3.
A: s(t+h) = 4(t+h)/(t+h+1). Difference quotient numerator: 4(t+h)(t+1) - 4t(t+h+1) = 4t² + 4t + 4ht + 4h - 4t² - 4ht - 4t = 4h. So the difference quotient is 4h / [h(t+h+1)(t+1)] = 4/[(t+h+1)(t+1)]. Derivative: s'(t) = 4/(t+1)². Velocity at t = 3: v(3) = 4/16 = 1/4 m/s.
Q: What algebraic technique do you use to simplify the difference quotient of f(x) = √(2x + 1)?
A: Multiply numerator and denominator by the conjugate, √(2(x+h)+1) + √(2x+1). This eliminates the square roots in the numerator via the identity (A - B)(A + B) = A² - B².
This material connects to differentiability and derivative graphs (Section 2.2, graphical side): the limit definition is the algebra behind the slopes you read off a graph. If the limit does not exist, the function is not differentiable there, which is exactly what corners, cusps and vertical tangents look like visually.
It also connects forward to the derivative rules (power rule, product rule, quotient rule, chain rule) in Sections 2.3 to 2.5. Those rules are shortcuts derived from this same limit definition. Once you have them, you will rarely use the limit definition in practice, but exams still test it directly to confirm you understand where the rules come from.
The tangent line equation reappears in linearisation and differentials (Section 3.9), where the tangent line is used as an approximation to the function near a point.
Related Terms / Search Tags
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