Adam - Limit Laws and Direct Substitution Property — Comprehensive Notes

  • Limit intuition and goals

    • When x approaches a, we ask what value the function f(x) gets arbitrarily close to. We denote this by the limit: lim⁡x→af(x)=ℓ.\lim_{x\to a} f(x)=\ell.

    • Some functions and numerics can mislead when using technology; in this class we rely on theorems (limit laws) to justify results.

    • The course builds toward the derivative; limit laws are building blocks for that development.

  • Setup for limit laws

    • Treat a constant c as a constant function: lim⁡x→ac=c.\lim_{x\to a} c=c.

    • Let f and g be functions with existing limits as x→a: lim⁡<em>x→af(x)=l,lim⁡</em>x→ag(x)=m.\lim<em>{x\to a} f(x)=l,\quad \lim</em>{x\to a} g(x)=m.

    • We assume a constant c and the limits in question exist for the laws below.

  • Limit laws (summary of rules mentioned)

    • Constant function (constant law): for any constant c,

    • lim⁡x→ac=c.\lim_{x\to a} c = c. (independent of a)

    • Addition law (sum):

    • lim⁡<em>x→a[f(x)+g(x)]=lim⁡</em>x→af(x)+lim⁡x→ag(x)=l+m.\lim<em>{x\to a} [f(x)+g(x)] = \lim</em>{x\to a} f(x) + \lim_{x\to a} g(x) = l+m.

    • Subtraction law (difference):

    • lim⁡x→a[f(x)−g(x)]=l−m.\lim_{x\to a} [f(x)-g(x)] = l-m.

    • Scalar/multiplication law: for a constant c,

    • lim⁡<em>x→a[c f(x)]=c lim⁡</em>x→af(x)=c l.\lim<em>{x\to a} [c\,f(x)] = c\,\lim</em>{x\to a} f(x) = c\,l.

    • Product law:

    • lim⁡x→a[f(x) g(x)]=(lim⁡f)(lim⁡g)=l m.\lim_{x\to a} [f(x)\,g(x)] = (\lim f)(\lim g) = l\,m.

    • Quotient law (with a caveat):

    • If lim⁡x→ag(x)≠0,\lim_{x\to a} g(x) \neq 0, then

    • lim⁡<em>x→af(x)g(x)=lim⁡</em>x→af(x)lim⁡x→ag(x)=lm.\lim<em>{x\to a} \frac{f(x)}{g(x)} = \frac{\lim</em>{x\to a} f(x)}{\lim_{x\to a} g(x)} = \frac{l}{m}.

    • Identity line law: the limit of the line xx as x→a is a:

    • lim⁡x→ax=a.\lim_{x\to a} x = a.

    • Power law (if f(x) has a limit L as x→a):

    • lim⁡x→a[f(x)]n=Ln.\lim_{x\to a} [f(x)]^{n} = L^{n}.

    • Special case: the power rule for x^n is a corollary of this when f(x)=x and L=a.

    • Root law (for nth roots):

    • lim⁡<em>x→af(x)n=lim⁡</em>x→af(x)n.\lim<em>{x\to a} \sqrt[n]{f(x)} = \sqrt[n]{\lim</em>{x\to a} f(x)}.

    • If n is even, a must satisfy a≥0 for the limit to be defined.

  • Direct Substitution Property (DSP)

    • If f is a polynomial or a rational function and a is in the domain of f, then

    • lim⁡x→af(x)=f(a).\lim_{x\to a} f(x) = f(a).

    • Note on domain: a symbol epsilon (\epsilon) is used in examples to denote "element of" the domain, i.e., a∈domain(f).

    • Example (DSP): lim_{x→2} (x^2+2)/(x-1) where the denominator is nonzero at x=2.

    • Since the denominator at a=2 is 1 (not zero), by DSP

    • lim⁡x→2x2+2x−1=(2)2+22−1=61=6.\lim_{x\to 2} \frac{x^2+2}{x-1} = \frac{(2)^2+2}{2-1} = \frac{6}{1} = 6.

  • When DSP cannot be used directly

    • If the denominator would be zero at a (or the function is not defined there), direct substitution fails.

    • In such cases, use algebraic manipulation (factoring, completing the square, rationalization, etc.) to simplify to a form where the limit can be evaluated.

    • Important caveat: after algebraic manipulation, the resulting limit equals the original limit, provided the manipulations are valid for x near a (and equal to f(x) for x≠a).

  • Examples worked step-by-step (illustrating limit laws and DSP)

    • Example 1: lim_{x→3} (x^2 + 2x - 1) / \sqrt[4]{x-2}

    • Check denominator at a=3: (3-2)^{1/4} = 1 ≠ 0, so no problem.

    • Top: by sum rule, lim = lim(x^2) + lim(2x) + lim(-1) = 9 + 6 + (-1) = 14.

    • Bottom: lim{x→3} (x-2)^{1/4} = [lim{x→3} (x-2)]^{1/4} = 1^{1/4} = 1.

    • Limit: 141=14.\frac{14}{1} = 14.

    • Example 2: lim_{x→2} (3x^2 - 2x - 1) / (x - 1)

    • Denominator at a=2 is 1 ≠ 0; can use DSP on the bottom; or substitute after evaluating the top limit.

    • Top: 3x^2 → 3(4) = 12; -2x → -4; -1 → -1; total = 12 - 4 - 1 = 7.

    • Denominator: lim_{x→2} (x - 1) = 1.

    • Limit: 71=7.\frac{7}{1} = 7.

    • Example 3: lim_{x→0} [\sqrt{x^2+4} - 2] / x^2

    • Direct substitution yields (√4 - 2)/0 → 0/0, so use algebraic manipulation via conjugate.

    • Multiply numerator and denominator by the conjugate: (\sqrt{x^2+4} + 2).

    • Numerator becomes (x^2 + 4) - 4 = x^2; denominator becomes x^2 [√(x^2+4) + 2].

    • Cancel x^2: limit becomes lim_{x→0} 1 / [√(x^2+4) + 2].

    • Direct substitution: 1 / [√(0+4) + 2] = 1 / (2 + 2) = 1/4.

    • Example 4: One-sided limits and piecewise function (to illustrate non-existence of a limit)

    • Piecewise function: f(x) = { x^2 + 1 if x < 1; (x-2)^2 if x ≥ 1 }.

    • Left-hand limit at a=1: lim_{x→1^-} f(x) = 1^2 + 1 = 2.

    • Right-hand limit at a=1: lim_{x→1^+} f(x) = (1-2)^2 = 1.

    • Since left and right limits are not equal, the limit as x→1 does not exist.

  • Theorem: existence of a two-sided limit via one-sided limits

    • The limit lim⁡x→af(x)=L\lim_{x\to a} f(x) = L exists if and only if the left-hand and right-hand limits both exist and equal L:

    • lim⁡<em>x→a−f(x)=Landlim⁡</em>x→a+f(x)=L.\lim<em>{x\to a^-} f(x) = L \quad\text{and}\quad \lim</em>{x\to a^+} f(x) = L.

    • This theorem helps diagnose limits that are difficult to approach from both sides (e.g., absolute value, square root, or piecewise definitions).

  • Practical problem-solving strategies highlighted

    • Before using limit laws, check the denominator at the target a to see if direct substitution is possible (DSP) or if division by zero would occur.

    • If the bottom is nonzero at a and both top and bottom are well-behaved, direct substitution is often fastest (DSP).

    • If the denominator goes to zero, employ factoring or algebraic manipulation to cancel problematic factors or to simplify the expression to a form where the limit can be evaluated.

    • Be mindful that f(x) and a manipulated g(x) might not be exactly the same for all x, but their limits as x→a agree when the manipulation is valid for x near a (x ≠ a if needed).

  • Worked set of problems (additional practice references mentioned in class)

    • Problem A: lim_{x→4} (x^2 - 4x) / (x^2 - 3x - 4)

    • Factor: top = x(x-4); bottom = (x-4)(x+1).

    • Cancel the common factor (x-4).

    • Remaining: lim_{x→4} [x / (x+1)].

    • Evaluate: 4 / (4+1) = 4/5.

    • Problem B: lim_{x→-1} (2x^2 + 3x + 1) / (x^2 - 2x - 3)

    • Factor both: numerator = (2x+1)(x+1); denominator = (x+1)(x-3).

    • Cancel (x+1) (valid for x ≠ -1).

    • Simplified form: (2x+1)/(x-3).

    • Evaluate at x→-1: (2(-1)+1)/(-1-3) = (-2+1)/(-4) = (-1)/(-4) = 1/4.

    • Problem C: lim_{x→0} [\sqrt{x^2+4} - 2] / x^2

    • This is the same as the conjugate example above, yielding 1/4 (as shown in Example 3).

  • Practical notes and classroom culture

    • Students are encouraged to discuss and help each other while solving problems. Collaboration is welcomed.

    • Avoid using L'Hôpital's rule until it has been covered in class.

    • The lecturer frames these exercises as tedious but foundational toward derivatives and more advanced limits.

  • Quick recap and bridge to next topic

    • We reviewed the basic limit laws (constant, sum, difference, scalar multiple, product, quotient, identity, power/root).

    • We reinforced the Direct Substitution Property for polynomials and rational functions.

    • We practiced algebraic manipulation when direct substitution fails, including factoring and rationalization via conjugates.

    • We introduced the two one-sided limits concept and its role in establishing the existence of a two-sided limit.

    • These tools prepare us for the concept and computation of the derivative in upcoming material.

  • Notation reminder

    • All mathematical expressions are presented in LaTeX formatting for clarity, e.g. lim⁡x→af(x)=ℓ\lim_{x\to a} f(x) = \ell, f(x)n\sqrt[n]{f(x)}, and so on.

  • End-of-session note

    • The lecturer expects to return to 2.3 and finish more problems tomorrow, reinforcing the same ideas with additional examples.