12.1Estimating Limits Numerically and Graphically
Estimate lim(x→c) f(x) from tables and graphs. Understand one-sided limits lim(x→c⁺) and lim(x→c⁻). Identify when limits do not exist (oscillation, unbounded behavior, jump discontinuity).
Limits are the foundational concept of calculus. Every derivative and integral is defined as a limit. Building intuition for limits numerically and graphically before the formal definition makes calculus far more approachable.
Essential Question
How can we describe the behavior of a function near a point — even when the function is not defined there?
A limit describes what value a function approaches as the input gets closer and closer to some value c. The function does not need to be defined at c for the limit to exist — we only care about behavior near c.
We write: lim(x→c) f(x) = L, which reads "the limit of f(x) as x approaches c equals L."
Two powerful strategies for estimating limits before using algebra are: (1) building a numerical table of values approaching c from both sides, and (2) reading the graph to see where the curve is heading.
Key Vocabulary
Limit
The value L that f(x) approaches as x gets arbitrarily close to c, written lim(x→c) f(x) = L.
Example: lim(x→2) (x²−4)/(x−2) = 4, even though the expression is undefined at x=2.
One-sided limit (left-hand)
lim(x→c⁻) f(x): the limit as x approaches c from values less than c (from the left).
Example: lim(x→3⁻) f(x) = 5 means f(x) → 5 as x approaches 3 through values like 2.9, 2.99, 2.999.
One-sided limit (right-hand)
lim(x→c⁺) f(x): the limit as x approaches c from values greater than c (from the right).
Example: lim(x→3⁺) f(x) = 7 means f(x) → 7 as x approaches 3 through values like 3.1, 3.01, 3.001.
Limit Does Not Exist (DNE)
A two-sided limit fails to exist when the left-hand and right-hand limits are unequal, the function oscillates without settling, or the function grows without bound.
Example: lim(x→0) sin(1/x) DNE because the function oscillates infinitely between −1 and 1.
Removable discontinuity (hole)
A point where a function is undefined (or defined differently) but the limit exists. The graph shows an open circle.
Example: f(x) = (x²−4)/(x−2) has a hole at x=2 because the factor (x−2) cancels.
Worked Examples
Estimate lim(x→2) (x²−4)/(x−2) using a table of values.
Note that x=2 makes the denominator zero, so f(2) is undefined. We cannot simply substitute.
Build a table approaching x=2 from both sides:
x = 1.9 → f(x) = (1.9²−4)/(1.9−2) = (3.61−4)/(−0.1) = (−0.39)/(−0.1) = 3.9
x = 1.99 → f(x) = (1.99²−4)/(1.99−2) = (3.9601−4)/(−0.01) = (−0.0399)/(−0.01) = 3.99
x = 1.999 → f(x) ≈ 3.999
x = 2.001 → f(x) ≈ 4.001
x = 2.01 → f(x) ≈ 4.01
x = 2.1 → f(x) = (2.1²−4)/(2.1−2) = (4.41−4)/(0.1) = 0.41/0.1 = 4.1
From both sides the values approach 4. We can verify: (x²−4)/(x−2) = (x+2)(x−2)/(x−2) = x+2 for x≠2, so the limit is 2+2=4.
Estimate lim(x→0) sin(x)/x using a table of values (x in radians).
x=0 makes the expression 0/0 (indeterminate). Build a table:
x = −0.1 → sin(−0.1)/(−0.1) ≈ (−0.09983)/(−0.1) ≈ 0.9983
x = −0.01 → sin(−0.01)/(−0.01) ≈ 0.99998
x = −0.001 → ≈ 0.9999998
x = 0.001 → ≈ 0.9999998
x = 0.01 → ≈ 0.99998
x = 0.1 → ≈ 0.9983
Values from both sides approach 1. This is a famous limit used throughout calculus and trigonometry.
A graph shows a function f with a removable discontinuity at x=1. The open circle is at (1, 2) and the filled dot is at (1, 4). What is lim(x→1) f(x)?
The filled dot at (1, 4) tells us f(1) = 4 — this is the actual function value.
The open circle at (1, 2) shows where the curve is heading from both sides.
The limit is determined by where the curve approaches, NOT by f(1).
From the left: the curve approaches y=2. From the right: the curve approaches y=2.
Since both one-sided limits equal 2, the two-sided limit exists and equals 2.
Let f(x) = { x+1 if x < 3 ; 2x−4 if x ≥ 3 }. Find lim(x→3⁻) f(x), lim(x→3⁺) f(x), and lim(x→3) f(x).
Left-hand limit: use the rule for x < 3, which is f(x) = x+1.
lim(x→3⁻) f(x) = lim(x→3⁻) (x+1) = 3+1 = 4
Right-hand limit: use the rule for x ≥ 3, which is f(x) = 2x−4.
lim(x→3⁺) f(x) = lim(x→3⁺) (2x−4) = 2(3)−4 = 6−4 = 2
For the two-sided limit to exist, both one-sided limits must be equal.
Left limit = 4, Right limit = 2. They are NOT equal.
Therefore the two-sided limit does not exist.
Explain why lim(x→0) sin(1/x) does not exist.
As x→0, the argument 1/x grows without bound.
sin(1/x) oscillates between −1 and +1 infinitely many times in any interval around 0.
Build a table: at x = 2/(π) ≈ 0.637, sin(1/x)=sin(π/2)=1.
At x = 2/(3π) ≈ 0.212, sin(1/x)=sin(3π/2)=−1.
At x = 2/(5π) ≈ 0.127, sin(1/x)=1 again.
The values never settle toward a single number — they keep bouncing between −1 and 1.
Since the function does not approach a single value L, the limit does not exist.
Guided Practice
Build a table of values and estimate lim(x→3) (x²−9)/(x−3). Use x = 2.9, 2.99, 2.999, 3.001, 3.01, 3.1.
Hint: Factor the numerator: x²−9 = (x+3)(x−3). After canceling (x−3), what simple expression remains? Evaluate that expression as x→3.
Estimate lim(x→0) (1−cos x)/x using a table with x = ±0.1, ±0.01, ±0.001 (x in radians).
Hint: Compute (1−cos(0.1))/0.1 ≈ (1−0.99500)/0.1 = 0.00500/0.1 = 0.0500. Do the same for smaller values. What number are all the outputs approaching?
A graph of f shows: for x < 2, the curve approaches y=5 from the left; for x > 2, the curve approaches y=5 from the right; but f(2) = 1 (filled dot). What is lim(x→2) f(x)? Does it equal f(2)?
Hint: Remember: the limit describes where the curve is heading, not the actual function value. Check whether the left-hand and right-hand limits agree.
Let g(x) = { 3x−2 if x < 4 ; x²−6 if x ≥ 4 }. Find lim(x→4⁻) g(x) and lim(x→4⁺) g(x). Does lim(x→4) g(x) exist?
Hint: For the left-hand limit, substitute x=4 into 3x−2. For the right-hand limit, substitute x=4 into x²−6. Compare the two results.
You are told lim(x→1⁻) f(x) = 7 and lim(x→1⁺) f(x) = 7. Does lim(x→1) f(x) exist? What is it? Does this tell you anything about f(1)?
Hint: The two-sided limit exists if and only if both one-sided limits exist and are equal. The value of f(1) is irrelevant to the limit — it could be 7, or 100, or undefined.
Common Mistakes
Common Mistakes
Assuming lim(x→c) f(x) = f(c) always. Students plug in x=c and call it the limit.
The limit describes behavior near c, not at c. Always check whether the function is continuous at c before substituting directly.
Concluding a limit exists because the function is defined at x=c.
A function can be defined at c (f(c) exists) but still have a limit that differs from f(c), or even no limit at all (jump discontinuity).
Using only one side of the table (e.g., only x values less than c) to estimate a two-sided limit.
Always approach c from BOTH sides. If the left and right outputs don't agree, the two-sided limit does not exist.
Saying lim(x→0) sin(1/x) = 0 because sin is bounded and 'small x makes it go to zero'.
The argument 1/x blows up as x→0, causing sin(1/x) to oscillate wildly. The limit DNE — never confuse boundedness with convergence.
Confusing lim(x→c⁻) f(x) and lim(x→c⁺) f(x) — mixing up which side is left vs. right.
The superscript '−' means approaching from the LEFT (values smaller than c). The superscript '+' means from the RIGHT (values larger than c).
Math Tips
Math Tips
When building a table, use at least 3 values on each side, getting progressively closer to c (e.g., 0.1, 0.01, 0.001). More decimal places = more confidence in your estimate.
If a table gives 0/0 at x=c, try factoring and canceling. The simplified expression usually reveals the limit directly.
On a graph, trace the curve with your finger from the left and from the right. Where do your fingers meet? That y-value is the limit (if they meet at the same height).
The two-sided limit exists if and only if: lim(x→c⁻) f(x) = lim(x→c⁺) f(x). This is the key theorem for all limit existence checks.
Three classic DNE scenarios to memorize: (1) jump discontinuity — left ≠ right; (2) oscillation — sin(1/x) type; (3) unbounded — vertical asymptote.
Interactive Practice — 5 Questions
A table of values shows f(x) approaching 6 from both sides as x→4, but f(4) = 10. What is lim(x→4) f(x)?
For a piecewise function, lim(x→5⁻) f(x) = 3 and lim(x→5⁺) f(x) = 7. What is lim(x→5) f(x)?
Which of the following best explains why lim(x→0) sin(1/x) does not exist?
A graph shows an open circle at (2, 5) and a filled dot at (2, 1). What is lim(x→2) f(x)?
Using a table, you find: as x→0 from both sides, (1−cos x)/x approaches 0. What does this tell you?