Limit Definition of the Derivative Practice Problems and Solutions

These limit definition of the derivative practice problems start with a line and a parabola, then move to cubes, fractions and square roots, and finish with tangent lines and limits you can recognize as derivatives. Try each one before you read the solution. The answers come right after the list.

The definition you use every time:

For a refresher on where it comes from, read the limit definition of the derivative guide.

Problems

Polynomials

1. f(x) = 5x - 3
2. f(x) = x^2
3. f(x) = 3x^2 - 4x + 1
4. f(x) = x^3

Fractions and roots

5. f(x) = \dfrac{1}{x}
6. f(x) = \sqrt{x}
7. f(x) = \dfrac{1}{x + 2}
8. f(x) = \sqrt{2x + 1}

At a point and tangent lines

9. Use the definition to find f'(3) for f(x) = x^2 + 2x.
10. Find the tangent line to f(x) = x^2 - 3x at x = 1, using the definition for the slope.

Recognize the derivative

11. \displaystyle\lim_{h\to 0}\frac{(2+h)^4 - 16}{h}
12. \displaystyle\lim_{h\to 0}\frac{\sin\!\left(\frac{\pi}{3} + h\right) - \sin\frac{\pi}{3}}{h}

Answers

Step-by-step solutions

1. f(x+h) = 5(x+h) - 3 = 5x + 5h - 3. Subtract f(x):

A line has the same slope everywhere, so f'(x) = 5.

2. Expand (x+h)^2 = x^2 + 2xh + h^2:

3. f(x+h) = 3(x^2 + 2xh + h^2) - 4x - 4h + 1. After subtracting f(x), only the terms with h remain:

4. Expand (x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3:

5. Combine the fractions over x(x+h):

6. Multiply top and bottom by the conjugate \sqrt{x+h} + \sqrt{x}:

As h\to 0 this becomes \frac{1}{2\sqrt{x}}.

7. Same idea as problem 5, with common denominator (x + h + 2)(x + 2):

8. Multiply by the conjugate. The top becomes (2x + 2h + 1) - (2x + 1) = 2h:

9. Use f'(3) = \lim_{h\to 0}\frac{f(3+h) - f(3)}{h}. Here f(3) = 15 and

10. First the slope at x = 1. f(1) = -2 and f(1 + h) = (1 + h)^2 - 3(1 + h) = -2 - h + h^2:

The tangent line goes through (1, -2) with slope -1:

11. This matches \frac{f(a + h) - f(a)}{h} with f(x) = x^4 and a = 2, since 2^4 = 16. So the limit is f'(2), and with the power rule, f'(x) = 4x^3 gives f'(2) = 32.

12. This matches f(x) = \sin x at a = \frac{\pi}{3}. The derivative of \sin x is \cos x, so the limit is \cos\frac{\pi}{3} = \frac{1}{2}.

Common mistakes on these problems

• Writing f(x+h) as f(x) + h. In problem 2, (x+h)^2 is x^2 + 2xh + h^2, not x^2 + h.
• Losing a sign when subtracting. Put f(x) in brackets before you subtract, especially in problem 3 where it has three terms.
• Letting h\to 0 too early. Plugging in h = 0 before cancelling gives \frac{0}{0}. Cancel the h first.
• Expanding the conjugate on the bottom. In problems 6 and 8, leave the bottom factored. You only need the top to simplify.

Keep practicing

These 12 problems use fixed numbers. For unlimited practice with new functions and instant feedback, try the limit definition lesson and the tangent lines lesson. Once the definition feels solid, the shortcut is the power rule: practice it with the power rule practice problems.