Difficulty: Introductory to Intermediate | Prerequisites: Algebra, basic trigonometry, familiarity with logarithmic and exponential functions
A limit describes the value a function approaches as its input approaches a particular point, even if the function is undefined there. Evaluating limits is the foundational skill behind derivatives and integrals: every major idea in calculus rests on it. The core techniques are direct substitution, algebraic simplification (factoring, conjugate multiplication), dominant-term analysis for limits at infinity, and L'Hôpital's Rule for indeterminate forms.
Limit
The value that f(x) approaches as x approaches a given number c. Written lim (x→c) f(x) = L.
In simple terms, this means: "What y-value does the curve get closer and closer to as x gets closer and closer to c?"
Indeterminate form
An expression such as 0/0 or ∞/∞ that arises from direct substitution and whose value cannot be determined without further work.
Think of it as a signal, not an answer. It tells you that the limit exists but you need another technique to find it.
Conjugate multiplication (rationalisation)
Multiplying numerator and denominator by the conjugate of a radical expression to eliminate the square root and simplify a 0/0 form.
In simple terms: if you see √(something) minus a number in the numerator, multiply top and bottom by √(something) plus that number.
Factoring and cancellation
Rewriting the numerator and denominator as products to cancel the common factor that causes the 0/0 form.
Think of it as removing the "hole" in the function so you can substitute directly.
Dominant-term analysis
For limits at ±∞, identifying the highest-degree term in the numerator and denominator and ignoring lower-order terms, because they become negligible.
In simple terms: as x gets enormous, only the biggest power of x matters.
L'Hôpital's Rule
If lim (x→c) f(x)/g(x) yields 0/0 or ±∞/∞, then lim (x→c) f(x)/g(x) = lim (x→c) f'(x)/g'(x), provided the latter limit exists.
Think of it as a shortcut: differentiate the top and bottom separately (not using the quotient rule) and try substitution again.
Use this when you see a square root in the numerator or denominator and direct substitution gives 0/0.
Worked example: lim (x→4) [√(x+5) − 3] / (x − 4)
Substituting x = 4 gives √9 − 3 = 0 in the numerator and 4 − 4 = 0 in the denominator, so you get 0/0.
Multiply numerator and denominator by the conjugate √(x+5) + 3.
The numerator becomes (x + 5) − 9 = x − 4 by the difference-of-squares identity.
Cancel the (x − 4) factor from numerator and denominator.
You are left with 1 / [√(x+5) + 3]. Substituting x = 4 gives 1 / (3 + 3) = 1/6.
Use this when both numerator and denominator are polynomials and direct substitution gives 0/0.
Worked example: lim (x→5/2) (4x² − 25) / (6x² − 7x − 20)
Substituting x = 5/2 gives 0/0.
Factor the numerator as a difference of squares: 4x² − 25 = (2x + 5)(2x − 5).
Factor the denominator: 6x² − 7x − 20 = (2x − 5)(3x + 4).
Cancel (2x − 5).
The limit becomes (2x + 5) / (3x + 4). Substituting x = 5/2 gives 10 / (23/2) = 20/23.
For rational expressions as x → ±∞, only the highest-power terms in the numerator and denominator determine the limit.
Worked example 3a: lim (x→ −∞) (3 − 2x² + 5x − √x) / (∛x + 4 − 3x + 8x²)
The numerator is dominated by −2x², and the denominator is dominated by 8x² (the cube root and √x grow more slowly than x²).
The limit behaves like −2x² / 8x² = −1/4.
Worked example 3b: lim (x→∞) (5 + x − 4x²) / (6 − 18x + 7xᵖⁱ)
The numerator's highest power is x². The denominator's highest power is xᵖⁱ (π ≈ 3.14).
Since π > 2, the denominator grows faster than the numerator.
The limit is 0.
Worked example 3c: lim (x→∞) (x + 2)(x³ − 64) / (16 − x³)
Expanding the numerator, the leading term is x⁴. The denominator's leading term is −x³.
The expression behaves like x⁴ / (−x³) = −x, which diverges.
The limit is −∞.
Use this when algebraic tricks do not simplify a 0/0 or ∞/∞ form, particularly when logarithms or trigonometric functions are involved.
Worked example: lim (x→0) [ln(1 − x) − sin x] / [1 − x − cos²x]
Substituting x = 0 gives (0 − 0) / (1 − 0 − 1) = 0/0.
Differentiate numerator: d/dx [ln(1 − x) − sin x] = −1/(1 − x) − cos x.
Differentiate denominator: d/dx [1 − x − cos²x] = −1 + 2 cos x sin x = −1 + sin 2x.
Substituting x = 0 gives (−1 − 1) / (−1 + 0) = −2 / −1 = 2.
\lim_{x \to c} f(x) = LFormal definition: for every ε > 0 there exists a δ > 0 such that if 0 < |x − c| < δ, then |f(x) − L| < ε.
\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}L'Hôpital's Rule (applies only when both sides produce 0/0 or ±∞/∞).
Difference of squares identity, used in conjugate multiplication:
(a - b)(a + b) = a^2 - b^2Dominant-term shortcut for rational functions at infinity:
If degree of numerator < degree of denominator, the limit is 0.
If degree of numerator = degree of denominator, the limit is the ratio of leading coefficients.
If degree of numerator > degree of denominator, the limit is ±∞.
Students often think the limit of f(x) as x → c equals f(c). It does not, necessarily. The function need not even be defined at c for the limit to exist.
Students often apply L'Hôpital's Rule when direct substitution already gives a finite, non-indeterminate answer. The rule only applies to 0/0 or ∞/∞ forms. If substitution works, use it.
Students often use L'Hôpital's Rule and differentiate the numerator and denominator using the quotient rule. That is wrong. You differentiate the numerator and denominator separately.
Students often assume that ∞/∞ = 1. It does not. ∞/∞ is indeterminate, meaning the actual limit could be any real number, 0, or ±∞ depending on the functions involved.
⚠️ Limits are tested on virtually every calculus exam. Expect at least one problem requiring conjugate multiplication and one requiring dominant-term analysis.
⚠️ L'Hôpital's Rule problems are common and often involve logarithms or trigonometric functions. Be prepared to apply it more than once if the first application still yields an indeterminate form.
⚠️ Exams frequently test whether you can identify which technique applies. The first step is always: try direct substitution. If you get a number, you are done. If you get 0/0, choose a technique.
⚠️ "Compare the degrees" is a fast path on limits at infinity. Know the three cases (degree numerator less than, equal to, or greater than degree denominator) cold.
True or false: If f(4) is undefined, then lim (x→4) f(x) does not exist.
False. The function need not be defined at a point for the limit to exist there.
Fill in the blank: If direct substitution into a limit gives 0/0, the form is called ______.
Indeterminate.
True or false: L'Hôpital's Rule says to apply the quotient rule to f(x)/g(x).
False. You differentiate f(x) and g(x) separately.
Fill in the blank: For a rational function, if the degree of the numerator equals the degree of the denominator, the limit as x → ∞ equals ______.
The ratio of the leading coefficients.
True or false: The limit of (x² − 9)/(x − 3) as x → 3 equals 6.
True. Factor to (x + 3)(x − 3)/(x − 3), cancel, and substitute.
Q: Evaluate lim (x→4) [√(x+5) − 3] / (x − 4).
A: Multiply by the conjugate. The answer is 1/6.
Q: Evaluate lim (x→5/2) (4x² − 25) / (6x² − 7x − 20).
A: Factor both as products containing (2x − 5), cancel, and substitute. The answer is 20/23.
Q: Evaluate lim (x→∞) (5 + x − 4x²) / (6 − 18x + 7xᵖⁱ). What is the key observation?
A: The denominator has a higher degree (π ≈ 3.14 > 2), so the limit is 0.
Q: Evaluate lim (x→0) [ln(1 − x) − sin x] / [1 − x − cos²x]. State which technique you use and why.
A: Direct substitution gives 0/0. Apply L'Hôpital's Rule: differentiate numerator and denominator separately. The answer is 2.
Q: Explain why lim (x→∞) (x + 2)(x³ − 64) / (16 − x³) diverges to −∞.
A: The numerator behaves like x⁴ and the denominator like −x³. The ratio behaves like −x, which diverges to −∞.
Limits connect directly to derivatives: the derivative is defined as a limit of the difference quotient. Every differentiation rule you will learn is proved using limits.
Limits also connect to integrals: the definite integral is defined as the limit of Riemann sums as the number of rectangles approaches infinity.
L'Hôpital's Rule itself requires you to differentiate, so comfort with basic derivative rules (covered in the Derivatives study notes) makes limit problems faster.
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