Source: Homework 03, Section 1.6 – Limit Laws | Calculus I, Michigan State University
Tags: one-sided limits, left-hand limit, right-hand limit, absolute value limits, Squeeze Theorem, Sandwich Theorem, Pinching Theorem, MTH 132, MSU
Difficulty: Introductory–Intermediate | Prerequisites: Limit laws and direct substitution (Section 1.6 Part 1), basic understanding of absolute value, familiarity with trigonometric functions.
Sometimes a function behaves differently as you approach a point from the left versus the right. One-sided limits capture that distinction, and they are the key to understanding piecewise functions and absolute-value expressions in limits. The Squeeze Theorem handles a different problem entirely: when you cannot simplify a function algebraically, but you can trap it between two simpler functions that share the same limit. Together, these tools round out the limit-evaluation toolkit you will use for the rest of the course.
If you are catching up, make sure you are comfortable with the basic limit laws and direct substitution first. This material builds directly on those skills.
One-sided limits tell you what a function does from just the left or just the right. The two-sided limit exists only when both one-sided limits agree. The Squeeze Theorem says: if a function is trapped between two others that converge to the same value, the trapped function must converge there too.
Left-hand limit (lim x→a⁻)
The value f(x) approaches as x gets close to a from values less than a (from the left on the number line). Think of it as "sneaking up on a from below."
Right-hand limit (lim x→a⁺)
The value f(x) approaches as x gets close to a from values greater than a (from the right). Think of it as "approaching a from above."
Two-sided limit
lim(x→a) f(x) exists if and only if the left-hand limit and the right-hand limit both exist and are equal. In simple terms, the function has to be heading to the same place from both directions.
Absolute value (in limits context)
|x – a| equals (x – a) when x > a, and –(x – a) when x < a. This piecewise definition is why absolute value expressions often require separate left and right analysis.
Squeeze Theorem (also called Sandwich Theorem or Pinching Theorem)
If g(x) ≤ f(x) ≤ h(x) for all x near a (except possibly at a), and lim(x→a) g(x) = lim(x→a) h(x) = L, then lim(x→a) f(x) = L. In simple terms, if two known functions pinch an unknown function to the same value, the unknown function must go there too.
The absolute value |x – a| is defined piecewise:
If x > a: |x – a| = x – a
If x < a: |x – a| = –(x – a) = a – x
This means any expression involving |x – a| / (x – a) will behave differently from the left and the right.
Worked example (Problem 11):
Evaluate lim(x→9) |x – 9| / (x – 9).
(a) From the left (x→9⁻): When x < 9, |x – 9| = –(x – 9). So the expression becomes –(x – 9)/(x – 9) = –1.
(b) From the right (x→9⁺): When x > 9, |x – 9| = (x – 9). So the expression becomes (x – 9)/(x – 9) = 1.
(c) Two-sided limit: The left-hand limit is –1 and the right-hand limit is 1. Since they disagree, lim(x→9) |x – 9|/(x – 9) does not exist.
Worked example (Problem 12):
Evaluate the one-sided limits of 6x(x – 5) / |x – 5|.
(a) From the left (x→5⁻): When x < 5, |x – 5| = –(x – 5). The expression becomes 6x(x – 5) / [–(x – 5)] = –6x. Substitute x = 5: –30.
(b) From the right (x→5⁺): When x > 5, |x – 5| = (x – 5). The expression becomes 6x(x – 5) / (x – 5) = 6x. Substitute x = 5: 30.
The key pattern: the (x – a) factors cancel, but the sign flips depending on which side you approach from.
The Squeeze Theorem requires three things:
A lower bound function g(x) ≤ f(x)
An upper bound function f(x) ≤ h(x)
Both bounds converge to the same limit: lim g(x) = lim h(x) = L
Then lim f(x) = L. The inequality does not need to hold at x = a itself, only near it.
Worked example (Problem 13):
Given 10x – 32 ≤ f(x) ≤ x² + 4x – 23, find lim(x→3) f(x).
Evaluate both bounds at x = 3:
Lower: 10(3) – 32 = 30 – 32 = –2
Upper: 9 + 12 – 23 = –2
Both bounds give –2. By the Squeeze Theorem, lim(x→3) f(x) = –2.
The method used: The Squeeze Theorem (not direct substitution, since we do not have a formula for f itself).
Worked example (Problem 14):
Given 2cos(2x) ≤ g(x) ≤ 2 + x³ for all x, find lim(x→0) g(x).
Evaluate both bounds at x = 0:
Lower: 2cos(0) = 2(1) = 2
Upper: 2 + 0³ = 2
Both bounds give 2. By the Squeeze Theorem, lim(x→0) g(x) = 2.
Note that the Squeeze Theorem is particularly useful when the function involves oscillating behaviour (like cosine) or when you are only given an inequality rather than an explicit formula.
Students often think that if the left-hand and right-hand limits both exist, the two-sided limit must exist. The two-sided limit exists only when both one-sided limits are equal. Existing but different is not enough.
When working with |x – a|/(x – a), students sometimes forget to consider both cases of the absolute value. Always split into x > a and x < a and analyse each separately.
Students sometimes try to use the Squeeze Theorem when the two bounding limits are different. The theorem only applies when both bounds converge to the same value.
A common error with the Squeeze Theorem is confusing it with direct substitution. If you are given an inequality for f(x) rather than a formula, direct substitution does not apply to f. You substitute into the bounds, not into f.
⚠️ Absolute value limit problems almost always require one-sided analysis. If you see |something| in a limit, split into cases.
⚠️ The Squeeze Theorem is the go-to method when you are given an inequality rather than an explicit function. Expect at least one problem asking you to identify which theorem you used.
⚠️ Remember that the two-sided limit does not exist when left and right limits disagree. "DNE" (does not exist) is a valid and expected answer on exams.
True or False: If lim(x→a⁻) f(x) = 3, then lim(x→a) f(x) = 3.
False. You also need the right-hand limit to equal 3.
Fill in the blank: When x < 7, |x – 7| = ___
–(x – 7), which equals 7 – x.
True or False: The Squeeze Theorem can be used even if the bounding functions have different limits.
False. The Squeeze Theorem requires both bounds to converge to the same value.
Fill in the blank: If –x² ≤ f(x) ≤ x² for all x, then lim(x→0) f(x) = ___
0, because both –x² and x² approach 0.
Q: Evaluate lim(x→2⁻) |x – 2| / (x – 2).
A: When x < 2, |x – 2| = –(x – 2). The expression becomes –(x – 2)/(x – 2) = –1.
Q: Evaluate lim(x→2⁺) |x – 2| / (x – 2).
A: When x > 2, |x – 2| = (x – 2). The expression becomes (x – 2)/(x – 2) = 1.
Q: Does lim(x→2) |x – 2| / (x – 2) exist?
A: No. The left-hand limit is –1 and the right-hand limit is 1. Since they are not equal, the two-sided limit does not exist.
Q: If 1 – x⁴ ≤ f(x) ≤ 1 + x⁴ for all x, find lim(x→0) f(x).
A: Both bounds approach 1 – 0 = 1 and 1 + 0 = 1 at x = 0. By the Squeeze Theorem, the limit is 1.
Q: Evaluate lim(x→3⁻) 4x(x – 3) / |x – 3|.
A: When x < 3, |x – 3| = –(x – 3). The expression becomes 4x(x – 3)/[–(x – 3)] = –4x. Substitute x = 3: –12.
One-sided limits connect directly to continuity (Section 1.8): a function is continuous at a point only if the two-sided limit exists and equals the function value, which requires both one-sided limits to agree. The Squeeze Theorem will reappear when you prove that lim(x→0) sin(x)/x = 1 in the trigonometric limits section, which is foundational for differentiating sine and cosine in Chapter 2. Understanding absolute-value behaviour in limits also prepares you for piecewise-defined functions and later for the formal definition of the derivative using one-sided limits.
one-sided limits, left-hand limit, right-hand limit, x approaches from left, x approaches from right, absolute value in limits, |x – a|/(x – a), piecewise limits, Squeeze Theorem, Sandwich Theorem, Pinching Theorem, trapped function, bounding functions, inequality limits, does not exist, DNE, two-sided limit, limit existence condition, MTH 132, Calculus I, MSU, Section 1.6, lim x→a⁻, lim x→a⁺