Learning Objectives
Select an appropriate method for a finite or one-sided limit.
Explain a discontinuity using values and limits.
Write a complete existence argument using the Intermediate Value Theorem.
Prerequisites: the first three lessons in this section. This review covers introductory limits and continuity. It does not yet cover every limit technique used later in calculus.
1. Choose a Method Before Calculating
Situation | First action |
|---|---|
Polynomial or nonzero rational denominator | Substitute the target input |
Substitution gives 0/0 | Factor or use a conjugate |
Nonzero numerator over a denominator tending to zero | Check signs on each side |
Piecewise boundary | Evaluate left limit, right limit, and value separately |
Proving an intermediate value exists | Check continuity and endpoint values |
Keep the question in view. A request for f(a), a limit at a, and continuity at a asks for three different things. Write the relevant definition before using algebra. For a two-sided limit, unequal one-sided results end the calculation: there is no single approached height.
2. Connect Simplification and Continuity
Worked Example: One Expression, Three Questions
Let f(x) = (x² + x − 6)/(x − 2) for x ≠ 2 and f(2) = 10. Find the limit at 2, decide continuity, and state a repair.
Factor the numerator as (x + 3)(x − 2). Near 2, f(x) = x + 3, so the limit is 5.
The defined value is 10, not 5. Thus f is discontinuous at 2. Changing f(2) to 5 repairs the removable discontinuity without changing any other value.
The cancelled factor explains the hole, but the domain restriction is still part of the original expression. This is why writing the condition x ≠ 2 alongside the simplification is useful rather than merely formal.
3. Explain Why a Claim Is Valid
Common Mistakes
A table alone does not prove that a limit exists; it samples only finitely many inputs.
A vertical asymptote is not a removable hole. Redefining one point cannot make unbounded nearby values continuous.
A sign change guarantees a root by the Intermediate Value Theorem only after continuity is established.
4. Mixed Practice
1. Evaluate lim(x → −2) (x² + 5x + 6)/(x + 2).
2. Evaluate lim(x → 0) (√(1 + x) − 1)/x.
3. Describe both one-sided limits of (x + 1)/(x − 1) at x = 1. Does a finite two-sided limit exist?
4. Let g(x) = x + k for x < 1 and g(x) = 3x for x ≥ 1. Find k for continuity at 1.
5. Show that x³ − x − 1 has at least one root between 1 and 2.
6. A learner claims that f(0) = 4 implies lim(x → 0) f(x) = 4. Give a counterexample and explain it.
Worked Solutions
1. Factor the numerator as (x + 2)(x + 3). For x ≠ −2, the quotient equals x + 3. The limit is 1.
2. Rationalizing gives 1/(√(1 + x) + 1) for x ≠ 0. Substitution now gives 1/2.
3. Near 1 the numerator is positive. The denominator is negative on the left and positive on the right. The left-hand limit is −∞ and the right-hand limit is +∞, so no finite two-sided limit exists.
4. The left-hand limit is 1 + k, while the right-hand limit and g(1) are 3. Hence 1 + k = 3 and k = 2.
5. A polynomial is continuous on [1, 2]. The endpoint values are −1 and 5, so 0 lies between them. The Intermediate Value Theorem guarantees a root in (1, 2).
6. Define f(x) = 0 for x ≠ 0 and f(0) = 4. All nearby outputs equal 0, so the limit is 0 although the function value is 4.
5. Summary
- Choose algebra based on the form obtained by substitution.
- A limit, a function value, and continuity must be checked separately.
- State the hypotheses of a theorem before using its conclusion.
- If a practice answer was wrong, identify whether the error was in method choice, algebra, or interpretation before trying again.