Difficulty: Introductory to Intermediate | Prerequisites: Rates of Change notes, basic factoring, trigonometric identities
Limits are the machinery that makes calculus work. Before you can differentiate or integrate, you need to evaluate what a function approaches as its input closes in on some value. Most of the techniques here are algebraic tricks for resolving expressions that initially look like 0/0 or ∞/∞. Master these and you have the toolkit for computing derivatives from the definition, and later for L'Hopital's Rule.
When direct substitution into a limit gives 0/0, you need a workaround: factor and cancel, multiply by the conjugate, use the standard limit sin(t)/t → 1, or apply the Squeeze Theorem. The goal every time is to rewrite the expression so the problematic division disappears.
Limit
The value that f(x) approaches as x approaches some value c, written lim (x → c) f(x) = L. Think of it as: what the function is heading towards, regardless of whether it ever arrives.
Indeterminate form
An expression like 0/0 or ∞/∞ that does not have a definite value on its own and requires further work. In simple terms, the function's numerator and denominator are both doing the same inconvenient thing, so you need algebra to untangle it.
Factoring (to resolve 0/0)
Rewriting numerator and denominator as products so that a common factor causing the 0/0 can be cancelled. Think of it as: identifying the shared troublemaker and removing it from both sides.
Conjugate multiplication
Multiplying the numerator and denominator by the conjugate of a radical expression (changing a + sign to − or vice versa) to eliminate a square root via the difference-of-squares identity. In simple terms, you are using (a − b)(a + b) = a² − b² to get rid of a stubborn radical.
Squeeze Theorem (Sandwich Theorem)
If g(x) ≤ f(x) ≤ h(x) near a point and g and h share the same limit L there, then f must also have limit L. Think of it as: if two known functions are squeezing an unknown function towards the same value, it has no choice but to go there.
Change of variables (substitution)
Replacing x with a new variable (e.g. t = 1/x) to transform a difficult limit into a more recognisable form. In simple terms, renaming the moving part so the expression looks like something you already know how to handle.
When direct substitution gives 0/0, try factoring both numerator and denominator. Cancel the common factor, then evaluate.
Worked example: lim (x → 1) (x² − 1) / (x² − 7x + 6)
Factor: (x − 1)(x + 1) / [(x − 6)(x − 1)]
Cancel (x − 1): (x + 1) / (x − 6)
Evaluate at x = 1: (1 + 1) / (1 − 6) = 2/(−5) = −2/5
Note: the simplified function (x + 1)/(x − 6) has a different domain from the original, but for limits that does not matter. Limits only care about behaviour near the point, not at it.
Useful when the expression contains a square root and factoring does not help.
Worked example: lim (t → 1) [√(3 + t) − 2] / (t − 1)
Multiply top and bottom by the conjugate √(3 + t) + 2.
Numerator becomes (3 + t) − 4 = t − 1 via the difference of squares.
Cancel (t − 1): left with 1 / [√(3 + t) + 2].
Evaluate at t = 1: 1 / (√4 + 2) = 1/4.
This identity is the workhorse for trig limits at 0. The strategy is to rewrite the expression so that the argument of sin matches the denominator.
Worked example: lim (x → 0) sin(6x) / (2x)
Rewrite as [sin(6x) / (6x)] · 3.
lim (x → 0) sin(6x)/(6x) = 1 by the standard limit.
The answer is 3 · 1 = 3.
Warning: sin(6x) ≠ 6 sin(x). This factoring is not valid, even though it happens to give the right answer here.
Nested example: lim (x → 0) sin(sin x) / x
Rewrite as [sin(sin x) / sin x] · [sin x / x].
Both factors approach 1 as x → 0.
The answer is 1 · 1 = 1.
When a limit at infinity does not fit standard forms, try substituting a new variable to turn it into a limit at 0 (or another known form).
Worked example: lim (x → −∞) x · sin(1/x)
As x → −∞, 1/x → 0⁻.
Set t = 1/x, so x = 1/t.
The limit becomes lim (t → 0⁻) (1/t) · sin(t) = lim (t → 0⁻) sin(t)/t = 1.
When you cannot simplify a function directly, try bounding it between two simpler functions that share the same limit.
Worked example: Suppose 1 − cos x ≤ f(x) ≤ x² for all x near 0 (except possibly at 0).
lim (x → 0) (1 − cos x) = 0.
lim (x → 0) x² = 0.
By the Squeeze Theorem, lim (x → 0) f(x) = 0.
This works for limits at infinity too. The key is that the bounding functions must converge to the same value.
Standard trig limit:
lim (t → 0) sin(t) / t = 1
Difference of squares (for conjugate method):
(a − b)(a + b) = a² − b²
Squeeze Theorem statement:
If g(x) ≤ f(x) ≤ h(x) near c (except possibly at c), and lim (x → c) g(x) = lim (x → c) h(x) = L, then lim (x → c) f(x) = L.
Change of variables for limits at infinity:
If x → ±∞, set t = 1/x so t → 0⁺ or 0⁻.
Limits underpin every derivative and integral you will ever compute. In engineering, limits describe what happens to a system as a parameter approaches a critical threshold, such as the stress on a beam approaching its yield point. In computer science, limits formalise what it means for an algorithm's running time to grow "like n²" as the input size gets large.
Students often assume that 0/0 means the limit does not exist. It does not: 0/0 is an indeterminate form, which means the limit could be any value (or none). You need more work to find it.
A common error with trig limits is writing sin(kx) = k sin(x). This is false. Use the identity lim sin(kx)/(kx) = 1 by matching the argument.
Students sometimes cancel factors that are zero at the limit point and worry that they have changed the function. For limits, this is fine: the limit only looks at behaviour near the point, never at it.
The Squeeze Theorem requires both bounding functions to have the same limit. If they converge to different values, the theorem tells you nothing.
⚠️ Expect at least one limit requiring factoring and cancellation of an (x − c) term.
⚠️ The sin(t)/t = 1 identity will likely appear. Practice rewriting expressions so the argument inside sin matches the denominator.
⚠️ Know when to use the conjugate trick: look for a square root in the numerator or denominator combined with a 0/0 form.
⚠️ The Squeeze Theorem may appear in a conceptual "show that this limit exists" question rather than a direct computation.
True or false: if direct substitution gives 0/0, the limit does not exist. False (it is indeterminate; further work is needed).
Fill in the blank: lim (x → 0) sin(5x)/(5x) = ______. 1.
True or false: when you cancel a common factor to evaluate a limit, you are changing the function's domain, but that is acceptable because limits only care about behaviour near the point. True.
Fill in the blank: to use the conjugate method on √(a) − b, multiply by ______ / ______. [√(a) + b] / [√(a) + b].
True or false: the Squeeze Theorem can be used for limits at infinity. True.
Q: Evaluate lim (x → 2) (x² − 4) / (x − 2).
A: Factor the numerator: (x − 2)(x + 2) / (x − 2) = x + 2. At x = 2: 2 + 2 = 4.
Q: Evaluate lim (t → 0) sin(3t) / (7t).
A: Rewrite as [sin(3t)/(3t)] · (3/7). The limit of sin(3t)/(3t) is 1, so the answer is 3/7.
Q: Evaluate lim (x → 4) [√x − 2] / (x − 4).
A: Multiply by the conjugate (√x + 2)/(√x + 2). Numerator becomes x − 4. Cancel: 1/(√x + 2). At x = 4: 1/(2 + 2) = 1/4.
Q: Suppose −x² ≤ f(x) ≤ x² for all x near 0. What is lim (x → 0) f(x)?
A: Both −x² and x² approach 0 as x → 0. By the Squeeze Theorem, lim (x → 0) f(x) = 0.
Q: Evaluate lim (x → 0) sin(sin x) / x. Show your reasoning.
A: Rewrite as [sin(sin x)/sin x] · [sin x / x]. As x → 0, sin x → 0, so the first factor → 1 by the standard limit; the second factor → 1 as well. The answer is 1.
These techniques are prerequisite for the formal definition of the limit (epsilon-delta proofs), for computing derivatives from the definition, and later for L'Hopital's Rule in Calculus II. The Squeeze Theorem reappears when you study convergence of sequences and series.
Limit calculation, indeterminate form 0/0, factoring limits, conjugate multiplication, rationalising the numerator, sin t over t, standard trig limit, Squeeze Theorem, Sandwich Theorem, change of variables in limits, substitution, MTH 132, Calculus I, Michigan State University, Sections 2.2-2.4