Connecting Multiple Representations of Limits: Study Guide (Calculus AB 1.9)

Course scope. This topic is shared by Calculus AB and Calculus BC. Everything here is examinable for both.

A note on notation

Limits are written in a compact form on this page:

lim (x → 3) f(x) means “the limit as x approaches 3 of f(x)”. lim (x → 3⁻) means x approaches 3 from the left (x < 3), and lim (x → 3⁺) means from the right (x > 3).

On paper, write the usual form with “x → 3” under “lim”.

One limit, four representations

Exam questions on limits give you information in four ways. The idea underneath is always the same: the limit as x → c of f(x) is the number that f(x) gets close to when x is close to c, but not equal to c.

Representation	How you read the limit	What can mislead you
Graph	Trace the curve towards x = c from the left and from the right. Note the height each side heads towards.	A filled dot at x = c shows f(c), not the limit. A small hole or spike can be invisible at the wrong scale.
Table	Read the outputs as the inputs close in on c from each side. Look for a trend, not a single entry.	Only a few inputs are shown. The function could behave differently between them.
Formula (analytic)	Substitute if allowed; otherwise rewrite (Topic 1.6) or use a theorem (Topic 1.8). For a piecewise formula, use the piece that applies on each side of c.	Using the piece for x = c itself instead of the pieces on either side.
Words (verbal)	Translate phrases into notation: “as x gets close to 5 from the right, P(x) gets close to 12” means lim (x → 5⁺) P(x) = 12.	Confusing “approaches” (a limit) with “equals at” (a function value).

Three ways a limit can fail to exist appear in every representation:

Way it fails	Example	What you see
Left limit ≠ right limit	(x − 2)/|x − 2| as x → 2	Graph jumps from −1 to 1; table values are −1 on the left and 1 on the right
Unbounded	1/x² as x → 0	Graph shoots upwards; table values grow without bound
Oscillating	sin(π/x) as x → 0	Graph swings between −1 and 1 faster and faster; no single trend

Reading a graph

Figure 1 shows a function f that this guide uses in the worked examples. Read it carefully: open circles mean “the curve gets close but this point is not on the graph”, filled dots mean “this point is on the graph”.

Two habits make graph questions safe:

1. Cover up x = c. Put a finger over the vertical line x = c. The limit is decided only by what you can still see either side of your finger.
2. Read each side separately. Write the left-hand and right-hand limits first, then decide about the two-sided limit.

Reading a table

A table gives a few sample inputs on each side of c. Read it like a sequence: are the outputs settling towards one number?

x	2.9	2.99	2.999	3.001	3.01	3.1
m(x)	5.8	5.98	5.998	6.002	6.02	6.2

The left-side outputs close in on 6 from below and the right-side outputs close in on 6 from above. A sensible estimate is lim (x → 3) m(x) = 6. Notice the words “sensible estimate”. The table does not show m(3), and it does not prove anything.

How a table can mislead. Take sin(π/x) near x = 0. At x = 0.1, 0.01 and 0.001, the value of π/x is 10π, 100π and 1000π, so sin(π/x) = 0 every time. A table built from those inputs suggests a limit of 0. But at x = 2/21 ≈ 0.095 the value is sin(10.5π) = 1, and at x = 2/23 ≈ 0.087 it is sin(11.5π) = −1. The function keeps swinging between −1 and 1 however close you get to 0, so the limit does not exist. The table hid this because of the inputs chosen.

How a graph can mislead. On a graphing screen, y = (x² − 4)/(x − 2) looks exactly like the line y = x + 2. The missing point at (2, 4) is one pixel wide or less, so you may not see it. The limit as x → 2 is still 4, but the function is undefined at x = 2. Scale can hide important behaviour; the formula reveals it.

Reading a formula and reading words

For a piecewise formula, the piece you use depends on the side. Suppose

• h(x) = x² − 1 for x < 2
• h(x) = 7 − x for x > 2
• h(2) = 5

From the left, use x² − 1: lim (x → 2⁻) h(x) = 4 − 1 = 3. From the right, use 7 − x: lim (x → 2⁺) h(x) = 5. The sides disagree, so lim (x → 2) h(x) does not exist. The value h(2) = 5 plays no part in either one-sided limit.

For words, translate one phrase at a time:

Sentence	Notation
“As t gets close to 4 from below, T(t) gets close to 20.”	lim (t → 4⁻) T(t) = 20
“Near x = 0, on both sides, g(x) gets close to −1.”	lim (x → 0) g(x) = −1
“As x approaches 2, the values of 1/(x − 2)² grow without bound.”	lim (x → 2) 1/(x − 2)² = ∞ (the limit does not exist as a real number)

Combining representations

Many questions give one function as a graph and another as a table or formula. The limit properties from Topic 1.5 still work, provided each limit you use exists:

• lim [f(x) ± g(x)] = lim f(x) ± lim g(x)
• lim [f(x) · g(x)] = lim f(x) · lim g(x)
• lim [f(x) / g(x)] = lim f(x) / lim g(x), but only when lim g(x) ≠ 0
• For a composite f(g(x)): find what g(x) approaches, say L, and from which side. Then use the limit of f as its input approaches L from that side. (This assumes g(x) ≠ L for x near c. If g(x) can equal L, the value f(L) matters too.)

If a two-sided limit does not exist, split into one-sided limits and work on each side.

Worked example 1: reading and combining limits from a graph

Question. Use the graph of f in Figure 1.

(a) Find lim (x → −1⁻) f(x), lim (x → −1⁺) f(x) and lim (x → −1) f(x). (b) Find lim (x → 1) f(x) and f(1). (c) Find lim (x → 1) [2f(x) − x²]. (d) Find lim (x → −1) [(x + 1) · f(x)].

Solution.

(a) From the left, the straight line rises towards the open circle at height 3, so lim (x → −1⁻) f(x) = 3. From the right, the parabola starts at the filled point (−1, 1), so lim (x → −1⁺) f(x) = 1. The one-sided limits are different, so lim (x → −1) f(x) does not exist.

(b) Both sides of the parabola rise towards the open circle at (1, 3). So lim (x → 1) f(x) = 3. The filled point says f(1) = 0. The limit and the function value are different numbers, and that is allowed.

(c) Both limits exist: lim (x → 1) f(x) = 3 and lim (x → 1) x² = 1. By the sum and constant-multiple properties,

lim (x → 1) [2f(x) − x²] = 2 × 3 − 1 = 5

A common error is to use f(1) = 0, which gives −1.

(d) The product rule needs both limits to exist, and lim (x → −1) f(x) does not. So work one side at a time.

• Left: (x + 1) → 0 and f(x) → 3, so the product → 0 × 3 = 0.
• Right: (x + 1) → 0 and f(x) → 1, so the product → 0 × 1 = 0.

Both one-sided limits equal 0, so lim (x → −1) [(x + 1) · f(x)] = 0.

Interpretation. Part (d) shows that an expression containing f can have a limit where f itself does not. Check each side.

Worked example 2: a graph and a table together

Question. The function g is defined for all x, and some of its values are shown below. Assume the trend in the table continues for all x close to 0, with g(x) > −1 for x ≠ 0.

x	−0.1	−0.01	−0.001	0.001	0.01	0.1
g(x)	−0.9	−0.99	−0.999	−0.999	−0.99	−0.9

Using this table and the graph of f in Figure 1, find:

(a) lim (x → 0) f(g(x)) (b) lim (x → 0) [g(x) · f(x + 1)] (c) lim (x → 0) [(f(x + 1) − 1) / g(x)]

Solution.

(a) Step 1: the inner function. From both sides, g(x) gets close to −1. So lim (x → 0) g(x) = −1.

Step 2: the side. Every value of g(x) is greater than −1. So the input to f approaches −1 from the right.

Step 3: the outer function. From the graph, lim (u → −1⁺) f(u) = 1. Therefore

lim (x → 0) f(g(x)) = 1

Saying “f has no limit at −1, so no answer” ignores the side: the input to f only arrives from the right.

(b) As x → 0, the input x + 1 → 1, from both sides. From Worked example 1(b), lim (u → 1) f(u) = 3, so lim (x → 0) f(x + 1) = 3. (The value f(1) = 0 is irrelevant, because x + 1 ≠ 1 when x ≠ 0.) Both limits exist, so by the product property

lim (x → 0) [g(x) · f(x + 1)] = (−1) × 3 = −3

Check. At x = 0.01, g(x) = −0.99 and f(1.01) is very close to 3 (the parabola is near its top), so the product is close to −3.

(c) The top has limit 3 − 1 = 2, using part (b). The bottom has limit −1, which is not 0, so the quotient property applies:

lim (x → 0) [(f(x + 1) − 1) / g(x)] = 2 / (−1) = −2

If the bottom limit had been 0, you could not divide the limits; you would need to rewrite (Topic 1.6) or check each side.

Worked example 3: from a formula to every other representation

Question. Let p(x) = x² + k for x < 3, p(x) = 2x + 5 for x > 3, and p(3) = 4, where k is a constant.

(a) Find the value of k for which lim (x → 3) p(x) exists, and state the limit. (b) For that k, describe the graph of p near x = 3 and write the limit as a sentence.

Solution.

(a) From the left: lim (x → 3⁻) (x² + k) = 9 + k. From the right: lim (x → 3⁺) (2x + 5) = 11. The two-sided limit exists only if 9 + k = 11, so k = 2, and then lim (x → 3) p(x) = 11.

(b) Graph: a parabola coming up to an open circle at (3, 11) from the left, a line leaving the same open circle to the right, and a separate filled point at (3, 4). Sentence: “As x gets close to 3 from either side, p(x) gets close to 11, even though p(3) = 4.”

Table check (k = 2): p(2.99) = 10.9401 and p(3.01) = 11.02. Both are close to 11.

Common misconceptions

• “The limit is where the filled dot is.” The filled dot gives f(c). Cover up x = c and look at the curve either side instead.
• “If the two-sided limit does not exist, nothing involving the function has a limit.” Products and composites can still have limits. Check each side (Worked example 1(d), Worked example 2(a)).
• Ignoring the side in a composite. For f(g(x)), it matters whether g(x) approaches its limit from above, from below or from both sides.
• Using f(L) instead of the limit of f at L. In Worked example 2(b), f(1) = 0 would give the wrong answer.
• Treating a table as proof. A table suggests a value. It can miss oscillation or a narrow spike between the sample inputs.
• Trusting one screen. A graphing window can hide a hole or a jump. Check the formula.
• Using the wrong piece of a piecewise function. The rule for x = c, or for the other side, does not affect a one-sided limit.
• Mixing up words. “Approaches” means a limit. “Equals at” or “is” means a function value.

Where this leads

Topic 1.10, Exploring Types of Discontinuities, uses exactly these readings. The hole at x = 1 in Figure 1 is a removable discontinuity, and the gap at x = −1 is a jump discontinuity. Later in Unit 1 you will meet unbounded behaviour again as vertical asymptotes. If the squeeze theorem from the previous topic felt shaky, revisit the Topic 1.8 study guide. Students in either course can return to the Calculus AB roadmap or the Calculus BC roadmap.

Try the practice questions now, then use the revision notes and the checklist to consolidate.