Limits: Quick Review

Key Concepts

Direct substitution is valid for continuous functions. If f is continuous at a, then lim<em>xaf(x)=f(a)\lim<em>{x\to a} f(x) = f(a). Polynomials are continuous everywhere, so limits equal the evaluation: lim</em>xap(x)=p(a)\lim</em>{x\to a} p(x) = p(a).
When the limit yields a form 0/0, algebraic manipulation (factoring, canceling common factors) is used to simplify before substitution.
When the limit involves infinity, analyze growth rates: if the dominant term grows without bound, the limit is ±∞.

Direct Substitution and Simple Finite Limits

If the function is simple and continuous, substitute the value. Examples: lim<em>x3x2=9\lim<em>{x\to 3} x^2 = 9, lim</em>x1(2x+1)=3\lim</em>{x\to 1} (2x+1) = 3, limx22x=4\lim_{x\to 2} 2x = 4.

Finite Polynomial Limits from the Transcript

From the transcript, certain explicit evaluations are correct:

  • limx0x2=0\lim_{x\to 0} x^2 = 0.
  • limx3x2=9\lim_{x\to 3} x^2 = 9.
  • limx2(2x22x+1)=5\lim_{x\to 2} (2x^2 - 2x + 1) = 5.

Limits at Infinity

For expressions that grow without bound as x grows, the limit is infinite: limx(2x2+1)=\lim_{x\to \infty} (2x^2 + 1) = \infty.

Quick Reference Card

  • If f is continuous at a: limxaf(x)=f(a)\lim_{x\to a} f(x) = f(a).
  • For polynomials: limxap(x)=p(a)\lim_{x\to a} p(x) = p(a).
  • For simple linear: limxa(mx+b)=ma+b\lim_{x\to a} (mx + b) = ma + b.
  • For rational expressions where the denominator tends to 0, check for factorization to cancel the zero if possible; otherwise, the limit may diverge.

Notes on the Transcript

Some parts of the provided transcript are garbled; the notes above capture the reliable, commonly applicable limit results and the explicit correct evaluations that were clear in the text.