The Limit Laws

If you torture data long enough, it will confess to anything or, in other words, there are different types of lies: white lies, seemingly small exaggerations and half-truths, damn lies, out-of-context information, and misleading statistics, #Anawim, justtothepoint.com.

Recall

Definition. A function f is a rule, relationship, or correspondence that assigns to each element of one set (x ∈ D), called the domain, exactly one element of a second set, called the range (y ∈ E).

The pair (x, y) is denoted as y = f(x). Typically, the sets D and E will be both the set of real numbers, ℝ. A mathematical function is like a black box that takes certain input values and generates corresponding output values (Figure E).

Very loosing speaking, a limit is the value to which a function grows close as the input get closer and closer to some other given value.

One would say that the limit of f, as x approaches a, is L, $\lim_{x \to a} f(x)=L$. Formally, for every real ε > 0, there exists a real δ > 0 such that for all real x, 0 < | x − a | < δ implies that | f(x) − L | < ε. In other words, f(x) gets closer and closer to L, f(x)∈ (L-ε, L+ε), as x moves closer and closer -approaching closer but never touching- to a (x ∈ (a-δ, a+δ), x≠a)) -Fig 1.a.-

Definition. Let f(x) be a function defined on an interval that contains x = a, except possibly at x = a, then we say that, $\lim_{x \to a} f(x) = L$ if

$\forall \epsilon>0, \exists \delta>0: 0<|x-a|<\delta, implies~ |f(x)-L|<\epsilon$

Or

$\forall \epsilon>0, \exists \delta>0: |f(x)-L|<\epsilon, whenever~ 0<|x-a|<\delta$

The Limit laws

Let f(x), g(x) be functions defined on an interval that contains x = a, except possibly at x = a, assume that L and M are real numbers such that $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$. Let c be a constant. Then, each of the following statements holds:

• Limit of a constant. The limit of a constant function is equal to the constant, $\lim_{x \to a} k = k$ where k is a constant, e.g., $lim_{x \to 12} 7 = 7$.
• Sum law for limits. It states that the limit of the sum of two functions equals the sum of the limits of both functions, that is, $\lim_{x \to a} (f(x)+g(x)) = lim_{x \to a} f(x) + lim_{x \to a} g(x) = L + M,$ e.g., $lim_{x \to 2}(2x+7) = lim_{x \to 2}(2x) + lim_{x \to 2}(7) = 4+7 = 11.$

Proof: Let $\epsilon>0$

$\exists \delta_1>0: 0<|x-a|<\delta_1, implies~ |f(x)-L|<\frac{\epsilon}{2}$

$\exists \delta_2>0: 0<|x-a|<\delta_2, implies~ |g(x)-M|<\frac{\epsilon}{2}$

Let’s choose $\delta = min (\delta_1, \delta_2).$

$\forall \epsilon>0, \exists \delta>0: 0<|x-a|<\delta$ ⇒[By the triangle inequality, |a+b|≤|a|+|b|] |f(x)+g(x)-L-M| ≤ |f(x)-L|+ |g(x)-M| $<\frac{\epsilon}{2} + <\frac{\epsilon}{2} = \epsilon$

• Difference law for limits. It states that the limit of the difference of two functions equals the difference of the limits of both functions, that is, $\lim_{x \to a} (f(x)-g(x)) = lim_{x \to a} f(x) - lim_{x \to a} g(x) = L - M,$ e.g., $lim_{x \to 2}(2x-7) = lim_{x \to 2}(2x) - lim_{x \to 2}(7) = 4-7 = -3.$
• Constant multiple law for limits. $\lim_{x \to a} c·f(x) = c·lim_{x \to a} f(x) = c·L,$ e.g., $\lim_{x \to 1} 7x^3 = 7·\lim_{x \to 1} x^3 = 7·1 = 7.$
• Product law for limits. It states that the limit of a product of two functions equals the product of the limits of both functions, that is, $\lim_{x \to a} (f(x)·g(x)) = lim_{x \to a} f(x) · lim_{x \to a} g(x) = L · M.$

$\lim_{x \to 2} x^2·(x+1) = \lim_{x \to 2} x^2·\lim_{x \to 2} x+1 = 4·3 = 12.$
$\lim_{x \to 2} (6x-x^3)·(x^2+3x-1) = \lim_{x \to 2} 6x-x^3·\lim_{x \to 2} x^2+3x-1 = 4·9 = 36.$ $\lim_{x \to 5} (2x^2-3x+4) = \lim_{x \to 5}(2x^2) - \lim_{x \to 5}(3x) + \lim_{x \to 5}(4) = 2·\lim_{x \to 5}(x^2) - 3·\lim_{x \to 5}(x) + \lim_{x \to 5}(4) = 2·5^2-3·5+4 = 39.$

Proof.

Let $\epsilon>0$

$\exists \delta>0: 0<|x-a|<\delta, implies~ |f(x)g(x)-LM|<\epsilon$

|f(x)g(x)-LM| = |f(x)g(x) -Lg(x) + Lg(x) -LM| = |g(x)(f(x)-L) + L(g(x)-M)| ≤[Triangle inequality] |g(x)(f(x)-L)| + |L(g(x)-M)| = |g(x)(f(x)-L)| + |L(g(x)-M)| = |g(x)||(f(x)-L)| + |L||(g(x)-M)|

This is the 🔑 tricky part:

1. $lim_{x \to a} g(x)$ = M ⇒ Let $\frac{ε}{2(|L|+1)}, \exists \delta_1>0: 0<|x-a|<\delta_1, implies~ |g(x)-M|<\frac{ε}{2(|L|+1)}.$
2. Let $ε = 1> 0, \exists \delta_2>0: 0<|x-a|<\delta_2, implies~ |g(x)-M|<1⇒|g(x)| = |g(x) -M +M| ≤ |g(x) -M| + |M|< 1 + |M|.$
3. $lim_{x \to a} f(x)$ = L ⇒ Let $\frac{ε}{2(|M|+1)}, \exists \delta_3>0: 0<|x-a|<\delta_3, implies~ |f(x)-L|<\frac{ε}{2(|M|+1)}.$

We choose δ = min{δ1, δ2, δ3}

|f(x)g(x)-LM| ≤ |g(x)||(f(x)-L)| + |L||(g(x)-M)| < $(1+|M|)\frac{ε}{2(|M|+1)}+|L|\frac{ε}{2(|L|+1)}<\frac{ε}{2}+|L|\frac{ε}{2|L|} = \frac{ε}{2}+\frac{ε}{2} = ε$∎

• Quotient law for limits. $\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)} = \frac{L}{M}$, e.g., $\lim_{x \to 2} \frac{x^2-1}{x-1} = \frac{\lim_{x \to 2} x^2-1}{\lim_{x \to 2} x-1} = \frac{3}{1} = 3.$
• Power law for limits. It states that the limit of the nth power of a function equals the nth power of the limit of the function, that is, $\lim_{x \to a} (f(x))^{n} = (\lim_{x \to a} f(x))^{n} = L^{n},$ e.g., $\lim_{x \to 5} (x+1)^{2} = (\lim_{x \to 5} x+1)^{2} = 6^{2} = 36.$
• Root law for limits. $\lim_{x \to a} \sqrt[n]{f(x)} = \sqrt[n]{\lim_{x \to a} f(x)} = \sqrt[n]{L}$ for all L if n is odd and for L≥0 if n is even. $\lim_{x \to 2} \sqrt{x^2+3x-1} = \sqrt{\lim_{x \to 2} x^2+3x-1} = \sqrt{9} = 3$ $\lim_{x \to 2} \sqrt[3]{\frac{5x+22}{2x}} = \sqrt[3]{\lim_{x \to 2} {\frac{5x+22}{2x}}} = \sqrt[3]{\frac{5·2+22}{2·2}} = \sqrt[3]{8} = 2.$
• Limit of a polynomial. $\lim_{x \to a} a_nx^n + a_{n-1}x^{n-1} + \ldots + a_2x^2 + a_1x + a_0 = a_n·a^n + a_{n-1}·a^{n-1} + \ldots + a_2·a^2 + a_1·a + a_0$, e.g., $\lim_{x \to 2} 3x^2-5x+7 = 3·2^2-5·2+7 = 12-10+7 = 9$.
• Limit of a composite function. The limit of a composition is the composition of the limit, provided the outside function is continuous at the limit of the inside function. $\lim_{x \to a} f ◦ g(x) = \lim_{x \to a} f(g(x)) = f(\lim_{x \to a} g(x))$ if f is continuous at $\lim_{x \to a} g(x).$ $\lim_{x \to 1} sin(\frac{π}{x+1}) = sin(\lim_{x \to 1} \frac{π}{x+1}) = sin(\frac{π}{2}) = 1.$ $\lim_{x \to 0} ln(cos(x)) = ln(\lim_{x \to 0} cos(x)) = ln(1)=$ [Recall e0 = 1] 0.
$\lim_{x \to ∞} cos(\frac{1}{x}) = cos(\lim_{x \to ∞} \frac{1}{x}) = cos(0) = 1.$ $\lim_{x \to 0⁺} x^x =$[$e^{ln(x^x)}=e^{x·ln(x)}$] $\lim_{x \to 0⁺} e^{x·ln(x)} = e^{\lim_{x \to 0⁺} x·ln(x)} = e^{\lim_{x \to 0⁺} \frac{ln(x)}{\frac{1}{x}}}=$[ L’Hospital’s Rule] $e^{\lim_{x \to 0⁺} \frac{\frac{1}{x}}{\frac{-1}{x^2}}} = e^{\lim_{x \to 0⁺} -x} = e^0 = 1.$

Bibliography

1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Calculus.
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn, and MathMajor.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. YouTube’s Andrew Misseldine: Calculus, College Algebra and Abstract Algebra.
8. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.
9. blackpenredpen.
Bitcoin donation