AP Calculus AB · Lesson 1

Finding limits: graphs, tables and algebra

Know Know it

The limit lim⁡x→af(x)=L\lim_{x \to a} f(x) = L means the outputs f(x)f(x) get as close to LL as you like when xx is close enough to aa, on both sides. The value f(a)f(a) plays no part: it can equal LL, differ from LL, or not exist at all.

A one-sided limit looks from one side only: lim⁡x→a−\lim_{x \to a^-} uses inputs less than aa, and lim⁡x→a+\lim_{x \to a^+} uses inputs greater than aa. The two-sided limit exists exactly when both one-sided limits exist and are equal.

On a graph, trace the curve toward x=ax = a from each side and read the height it heads for. An open circle marks a missing point; a filled dot marks the actual value f(a)f(a). In a table, read the outputs for inputs closing in on aa from each side. A table only suggests a limit; it cannot prove one.

Worked example Easy

The function ff is defined for −1≤x≤7-1 \le x \le 7. Its graph is shown.
The graph of y = f(x)−2−1123456782468100xy(2, 5)(2, 2)(4, 3)(4, 6)(-1, 2)(7, 9)
The graph of y=f(x)y = f(x) Three line segments. The first rises from (-1, 2) to an open circle at (2, 5). A filled dot sits alone at (2, 2). The second segment falls from the open circle at (2, 5) to an open circle at (4, 3). The third segment starts at a filled dot at (4, 6) and rises to (7, 9).

Based on the graph, what is lim⁡x→2f(x)\lim_{x \to 2} f(x)?

  1. 22
  2. 3.53.5
  3. 55Answer
  4. The limit does not exist

How to solve it

  1. Trace the graph toward x=2x = 2 from the left: it heads for the open circle at height 5.
  2. Trace it from the right: it heads for the same open circle.
  3. Both sides agree, so the limit is 5. The dot at (2, 2) is f(2)f(2), which does not matter for the limit.

Why each choice is right or wrong

  • A. Incorrect. 2 is the height of the filled dot, which is f(2)f(2), but a limit ignores the value at x=2x = 2 itself.
  • B. Incorrect. 3.5 averages the dot's height 2 and the curve's height 5, but a limit never averages anything.
  • C. Correct. From the left the graph climbs toward the open circle at (2, 5), and from the right it comes down toward the same circle, so both sides head for 5.
  • D. Incorrect. Saying the limit does not exist would need the two sides to head for different heights, but both approach 5.

Worked example Medium

The function ff is defined for −1≤x≤7-1 \le x \le 7. Its graph is shown.
The graph of y = f(x)−2−1123456782468100xy(2, 5)(2, 2)(4, 3)(4, 6)(-1, 2)(7, 9)
The graph of y=f(x)y = f(x) Three line segments. The first rises from (-1, 2) to an open circle at (2, 5). A filled dot sits alone at (2, 2). The second segment falls from the open circle at (2, 5) to an open circle at (4, 3). The third segment starts at a filled dot at (4, 6) and rises to (7, 9).

Which statement about ff near x=4x = 4 is true?

  1. lim⁡x→4f(x)=6\lim_{x \to 4} f(x) = 6, because f(4)=6f(4) = 6
  2. lim⁡x→4−f(x)=3\lim_{x \to 4^-} f(x) = 3, and lim⁡x→4f(x)\lim_{x \to 4} f(x) does not existAnswer
  3. lim⁡x→4+f(x)=3\lim_{x \to 4^+} f(x) = 3, and lim⁡x→4f(x)=3\lim_{x \to 4} f(x) = 3
  4. lim⁡x→4f(x)=4.5\lim_{x \to 4} f(x) = 4.5, the average of the two one-sided limits

How to solve it

  1. Left of 4, the middle segment falls toward the open circle at (4, 3), so the left-hand limit is 3.
  2. Right of 4, the last segment starts at (4, 6), so the right-hand limit is 6.
  3. The one-sided limits differ, so lim⁡x→4f(x)\lim_{x \to 4} f(x) does not exist, even though f(4)=6f(4) = 6 is defined.

Why each choice is right or wrong

  • A. Incorrect. f(4)=6f(4) = 6 is the filled dot, but the value at 4 does not set the limit, and the left side heads for 3, not 6.
  • B. Correct. From the left the graph falls toward the open circle at (4, 3); from the right it starts at height 6. The one-sided limits 3 and 6 differ, so the two-sided limit does not exist.
  • C. Incorrect. The right-hand limit at 4 is 6, from the segment that starts at (4, 6), not 3, so this statement gets the right side wrong.
  • D. Incorrect. Averaging 3 and 6 to get 4.5 is not a rule; when the one-sided limits differ, the two-sided limit does not exist.

Explain Explain it

Task verbs: explain, calculate

With a formula, try direct substitution first. Polynomials, roots, exponentials, logarithms, sine and cosine, and rational functions are continuous on their domains, so at an input in the domain the limit is just the value.

If substitution into a quotient gives 00\frac{0}{0}, the form is indeterminate: the algebra is unfinished. Factor and cancel, multiply by a conjugate, or combine fractions, then substitute again. If it gives a nonzero number over 0, the outputs grow without bound near that input and the limit is not a finite number.

Two more tools: lim⁡x→0sin⁡xx=1\lim_{x \to 0}\frac{\sin x}{x} = 1 (radians), and the squeeze theorem: if g(x)≤f(x)≤h(x)g(x) \le f(x) \le h(x) near aa and gg and hh have the same limit LL at aa, then so does ff.

Trap 0 over 0 means keep working

Getting 00\frac{0}{0} by substitution does not mean the limit is 0, or 1, or that it does not exist. It means you must simplify and substitute again. For example, x2−1x−1\frac{x^2 - 1}{x - 1} gives 00\frac{0}{0} at 1, yet it equals x+1x + 1 for x≠1x \ne 1, so its limit there is 2.

Worked example Hard

What is lim⁡x→3x+6−3x−3\lim_{x \to 3}\frac{\sqrt{x + 6} - 3}{x - 3}?

  1. 00
  2. 16\frac{1}{6}Answer
  3. 13\frac{1}{3}
  4. The limit does not exist

How to solve it

  1. Substitution gives 00\frac{0}{0}, so rationalize: multiply top and bottom by x+6+3\sqrt{x + 6} + 3.
  2. The top becomes (x+6)−9=x−3(x + 6) - 9 = x - 3, which cancels with the bottom for x≠3x \ne 3.
  3. What is left is 1x+6+3\frac{1}{\sqrt{x + 6} + 3}, and substituting 3 gives 13+3=16\frac{1}{3 + 3} = \frac{1}{6}.

Why each choice is right or wrong

  • A. Incorrect. 0 comes from looking only at the top, which goes to 0; the bottom also goes to 0, so the quotient needs work.
  • B. Correct. Multiplying by the conjugate gives (x+6)−9(x−3)(x+6+3)=1x+6+3\frac{(x + 6) - 9}{(x - 3)(\sqrt{x + 6} + 3)} = \frac{1}{\sqrt{x + 6} + 3}, which is 16\frac{1}{6} at x=3x = 3.
  • C. Incorrect. 13\frac{1}{3} divides by x+6\sqrt{x + 6} alone, dropping the +3+3 that comes with the conjugate.
  • D. Incorrect. A 00\frac{0}{0} form does not mean the limit fails to exist; after rationalizing, the limit is a finite number.

Explain Your turn

A table gives values of a function hh near x=5x = 5: h(4.9)=7.79h(4.9) = 7.79, h(4.99)=7.979h(4.99) = 7.979, h(4.999)=7.9979h(4.999) = 7.9979, h(5.001)=8.0021h(5.001) = 8.0021, h(5.01)=8.021h(5.01) = 8.021 and h(5.1)=8.21h(5.1) = 8.21. You are also told that h(5)=2h(5) = 2. Say what the table suggests lim⁡x→5h(x)\lim_{x \to 5} h(x) is, and explain why the value h(5)=2h(5) = 2 does not change that answer.

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Compare with a model answer

A full answer does this

  • Says the table suggests the limit is 8
  • Explains that a limit depends on inputs near 5, not on the value at 5

Model answer

From the left the outputs climb 7.79, 7.979, 7.9979, and from the right they fall 8.21, 8.021, 8.0021, so both sides close in on 8 and the table suggests lim⁡x→5h(x)=8\lim_{x \to 5} h(x) = 8. A limit describes h(x)h(x) for inputs close to 5 but not equal to 5, so the value h(5)=2h(5) = 2 has no effect on it. It only shows that hh is not continuous at 5.

Apply Apply it

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