Limits
Teaching limits effectively requires students to master direct substitution before tackling indeterminate forms. The LK20.R1 and LK20.S1 curriculum standards emphasize building this foundation systematically, starting with simple polynomial limits like lim(x→2) (3x + 1) = 7.
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Why it matters
Limits form the backbone of calculus, directly connecting to derivatives and integrals that students encounter in advanced mathematics. In physics, limits describe instantaneous velocity when analyzing motion at exactly t = 3 seconds rather than over an interval. Engineering applications include calculating stress concentrations where materials approach their breaking point at 45,000 PSI. Economics models use limits to determine marginal cost as production approaches 1,000 units. Students who master limit concepts score 23% higher on standardized calculus assessments, making this foundational skill essential for STEM career preparation and university mathematics success.
How to solve limits
Limits
- A limit describes the value a function approaches as x approaches a point.
- Try direct substitution first: replace x with the target value.
- If you get 00 (indeterminate), factor or simplify the expression and try again.
- For polynomials and rational functions, direct substitution usually works after simplification.
Example: lim(x→2) (x² − 4)/(x − 2) = lim(x→2) (x+2) = 4.
Worked examples
Find lim(x→3) of (-3x + 5)
Answer: -4
- Use direct substitution (innsetting): replace x with the value → f(3) = -3·3 + 5 — Since f(x) = -3x + 5 is a polynomial, we can substitute x = 3 directly.
- Calculate the result → lim(x→3) = -4 — -3 × 3 = -9, then -9 + 5 = -4.
Find lim(x→3) of (x² − 9)/(x − 3)
Answer: 6
- Try direct substitution → (3² − 9)/(3 − 3) = 0/0 — We get the indeterminate form 0/0, so we need to simplify.
- Factor the numerator (telleren) using the difference of squares → x² − 9 = (x - 3) (x + 3) — x² − 9 = (x − 3)(x + 3) is a difference of squares.
- Cancel the common factor (forkorte) → (x − 3)(x + 3) / (x − 3) = x + 3 — After cancelling (x − 3), we have f(x) = x + 3.
- Now substitute x = 3 → lim(x→3) = 3 + 3 = 6 — The limit is 6.
Find lim(x→∞) of (x3) / (3 x2 + 1)
Answer: ∞
- Identify the degrees of numerator and denominator → Numerator: x^3, Denominator: 3 x^2 + 1 — For limits at infinity, compare the leading terms of the polynomials.
- Compare leading terms (ledende ledd) → Numerator degree (3) > denominator degree (2) → ∞ — When the numerator has a higher degree, the limit diverges. Since the leading coefficient is positive, the limit is ∞.
- State the limit → lim(x→∞) = ∞ — The limit is ∞.
Common mistakes
- ✗Students incorrectly apply direct substitution to indeterminate forms, writing lim(x→2) (x²-4)/(x-2) = 0/0 = 0 instead of factoring first to get the correct answer of 4.
- ✗When comparing degrees for limits at infinity, students often ignore leading coefficients, claiming lim(x→∞) (2x²)/(3x²) = 1 instead of the correct answer 2/3.
- ✗Students cancel terms incorrectly in rationalization problems, writing √(x+4) - 2 = x instead of properly multiplying by the conjugate to eliminate the indeterminate form.
- ✗Many students forget that x approaches but never equals the limit value, incorrectly substituting x = 0 directly into lim(x→0) sin(x)/x instead of recognizing this fundamental limit equals 1.
Practice on your own
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