Limit Laws Using Graphs — Detailed Study Notes

Context / Setup

  • Two functions were supplied graphically: f(x) and g(x).
    • Exact algebraic formulas were NOT given; all numerical limits are read directly from the plotted y-values.
    • Points of interest: x=0,  1,  2x = 0,\;1,\;2.
  • Task: Evaluate six limits (labeled a–f) using limit laws.
  • Requirement: State “does not exist” (DNE) whenever the two-sided limit fails, and explain why.

Essential Limit Laws Recalled in the Video

  • Sum/Difference Law
    lim⁡<em>x→a [f(x)±g(x)]=lim⁡</em>x→af(x)±lim⁡x→ag(x)\lim<em>{x\to a}\,[f(x) \pm g(x)] = \lim</em>{x\to a}f(x) \pm \lim_{x\to a}g(x) (if both one-sided limits exist)
  • Product Law
    lim⁡<em>x→a[f(x) g(x)]=(lim⁡</em>x→af(x))(lim⁡x→ag(x))\lim<em>{x\to a}[f(x)\,g(x)] = \Bigl(\lim</em>{x\to a}f(x)\Bigr)\Bigl(\lim_{x\to a}g(x)\Bigr)
  • Quotient Law
    lim⁡<em>x→af(x)g(x)=lim⁡</em>x→af(x)lim⁡x→ag(x)\lim<em>{x\to a}\frac{f(x)}{g(x)} = \frac{\lim</em>{x\to a}f(x)}{\lim_{x\to a}g(x)} provided the denominator limit exists and is non-zero.
  • Power / Polynomial Law
    lim⁡x→axn=an\lim_{x\to a}x^n = a^n (continuous everywhere).
  • Constant Law
    lim⁡x→ac=c\lim_{x\to a}c = c
  • Root Law
    lim⁡<em>x→a h(x) n= lim⁡</em>x→ah(x) n\lim<em>{x\to a}\sqrt[n]{\,h(x)\,} = \sqrt[n]{\,\lim</em>{x\to a}h(x)\,} when the inside limit is positive (or otherwise lies in the domain of the root).
  • Continuity Shortcut
    • If a function is continuous at x=ax=a, the limit equals the function value: lim⁡x→af(x)=f(a)\lim_{x\to a}f(x)=f(a).

Graph-Based Numerical Limits Extracted Before Working Examples

Target xxf(x)f(x) (approach)g(x)g(x) (approach)Notes
0000≈1.4\approx1.4f crosses the origin; g sits slightly above 1.
1111LEFT (≈2) vs. RIGHT (≈0) → mismatch → limit DNEJump discontinuity in g.
222200f is continuous there; g touches the x-axis.

(The instructor constantly references these values while applying the laws.)


Worked Examples

(a) lim⁡x→2[f(x)+g(x)]\displaystyle \lim_{x\to2}[f(x)+g(x)]

  • Split via Sum Law: lim⁡<em>x→2f(x)+lim⁡</em>x→2g(x).\lim<em>{x\to2}f(x)+\lim</em>{x\to2}g(x).
  • From the graph: lim⁡<em>x→2f(x)=2,\lim<em>{x\to2}f(x)=2, lim⁡</em>x→2g(x)=0.\lim</em>{x\to2}g(x)=0.
  • Result: 2+0=2.2+0 = 2.
  • Comment: Both functions are continuous at x=2x=2, so direct substitution was valid.

(b) lim⁡x→1[f(x)+g(x)]\displaystyle \lim_{x\to1}[f(x)+g(x)]

  • Apply Sum Law again, but need each individual limit.
    • lim⁡<em>x→1f(x)=1\lim<em>{x\to1}f(x)=1 (f continuous). • lim⁡</em>x→1g(x)\lim</em>{x\to1}g(x) does not exist – left-hand limit ≈2\approx2, right-hand ≈0\approx0.
  • Because one addend’s limit fails, the entire sum’s limit does not exist (DNE).
  • Explanation provided: “two sides don’t match up.”

(c) lim⁡x→0[f(x) g(x)]\displaystyle \lim_{x\to0}[f(x)\,g(x)]

  • Product Law: (lim⁡<em>x→0f(x))(lim⁡</em>x→0g(x)).\bigl(\lim<em>{x\to0}f(x)\bigr)\bigl(\lim</em>{x\to0}g(x)\bigr).
  • Values: lim⁡<em>x→0f(x)=0,\lim<em>{x\to0}f(x)=0, lim⁡</em>x→0g(x)≈1.4.\lim</em>{x\to0}g(x)\approx1.4.
  • Multiplying: 0×1.4=0.0\times1.4 = 0.
  • Insight: Any finite number times zero gives a zero limit.

(d) lim⁡x→1f(x)g(x)\displaystyle \lim_{x\to1}\frac{f(x)}{g(x)}

  • Quotient Law requires both numerator & denominator limits.
    • Numerator: lim⁡<em>x→1f(x)=1.\lim<em>{x\to1}f(x)=1. • Denominator: lim⁡</em>x→1g(x)\lim</em>{x\to1}g(x) DNE (same jump issue as part b).
  • Because the denominator limit fails, the quotient’s limit does not exist.
  • Instructor’s phrase: “This limit isn’t going to exist either because the limit in the denominator doesn’t exist.”

(e) lim⁡x→2[x3 f(x)]\displaystyle \lim_{x\to2}[x^3\,f(x)]

  • View it as a product x3⋅f(x)x^3 \cdot f(x).
  • Product Law ⇒ (lim⁡<em>x→2x3)(lim⁡</em>x→2f(x)).\bigl(\lim<em>{x\to2}x^3\bigr)\bigl(\lim</em>{x\to2}f(x)\bigr).
    1. Power Law: lim⁡x→2x3=23=8.\lim_{x\to2}x^3 = 2^3 = 8.
    2. From the graph: lim⁡x→2f(x)=2.\lim_{x\to2}f(x)=2.
  • Multiply: 8×2=16.8\times2 = 16.
  • Note: Demonstrates combining “basic” polynomial limit with a graph-read limit.

(f) lim⁡x→1 3+f(x) \displaystyle \lim_{x\to1}\sqrt{\,3+f(x)\,}

  1. Root Law lets us pull the limit inside:
    = lim⁡x→1[3+f(x)] .= \sqrt{\,\lim_{x\to1}[3+f(x)]\,}.
  2. Sum Law inside the radical:
    = lim⁡<em>x→13+lim⁡</em>x→1f(x) .= \sqrt{\,\lim<em>{x\to1}3 + \lim</em>{x\to1}f(x)\,}.
  3. Constant Law: lim⁡x→13=3.\lim_{x\to1}3 = 3.
  4. From graph: lim⁡x→1f(x)=1.\lim_{x\to1}f(x)=1.
  5. Combine inside: 3+1=4.3 + 1 = 4.
  6. Evaluate root: 4=2.\sqrt{4} = 2.
  • Valid because 3+f(x)3+f(x) stays positive near x=1x=1, preserving domain of the square root.

Conceptual & Pedagogical Highlights

  • Graphical intuition is often faster than algebra for limit evaluation—provided discontinuities are recognized.
  • Sum, product, and quotient laws all require the individual component limits to exist first; one bad component kills the combined limit.
  • The Power Law and Constant Law are examples of continuity of basic elementary functions.
  • Root Law extends continuity to radicals, but only where the interior expression remains within the radical’s domain.
  • When using the quotient law, always check that the denominator limit is non-zero as well as existing; here, the issue was total non-existence.
  • Mixing multiple laws (as in parts e & f) shows how composite expressions can be broken apart systematically.

Quick Reference / Cheat Sheet

  • Continuous at a point? → Evaluate directly.
  • Jump in one side vs. the other? → Limit fails.
  • Product includes a factor whose limit is 0? → Overall limit is 0 (provided the other factor is finite).
  • Quotient denominator limit DNE or =0? → Combined limit DNE (or ±∞ if one-sided analysis applies).
  • Root of a limit vs. limit of a root? → Interchangeable if the inside stays in the domain.

Final Numerical Answers Recap

  • (a) 22
  • (b) DNE
  • (c) 00
  • (d) DNE
  • (e) 1616
  • (f) 22

These notes encapsulate the logic, laws, and numeric conclusions the instructor walked through in the video.