5.1.1 Definition & Notation 🔧
Determinations of specific values in a function, even where values don't exist.
$$\frac{x^2-4}{x-2}\medspace→\medspace\frac{2^2-4}{2-2}=\frac{0}{0}$$
| $$\lim\limits_{x→a^+}f(x)$$ |
The limit as a function approaches a point from the right |
| $$\lim\limits_{x→a^-}f(x)$$ |
The limit as a function approaches a point from the left |
| $$\lim\limits_{x→a}f(x)$$ |
The limit as a function approaches a point from the both sides |
5.1.2 Continuity
A function $f(x)$ is continuous at point $a$ if the limit exists and $f(a)$ is defined;
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The limit exists when $\lim_{{x→a}^-}f(x)=\lim_{{x→a}^+}f(x)=\lim_{{x→a}}f(x)$
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There is no change in functionality, meaning there are no corners at $f(a)$
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$\lim_{{x→a}}f(x)=f(a)$
5.1.3 Arithmetic Properties
The limit of a constant function is equal to the constant
$$\lim\limits_{x→a} c=c$$
The limit of a constant multiple of a function equals the product of the constant with the limit of the function
$$\lim\limits_{x→a} c·f(x)=c·\lim\limits_{x→a} f(x)$$
The limit of sums is the sum of the limits
$$\lim\limits_{x→a} \big(f(x)\pm g(x)\big)=\lim\limits_{x→a} f(x)\pm \lim\limits_{x→a} g(x)$$
The limit of products is the product of the limits
$$\lim\limits_{x→a} \big(f(x)·g(x)^{\pm 1}\big)=\lim\limits_{x→a} f(x)· \lim\limits_{x→a} g(x)^{\pm 1}$$
5.1.4 Limits of Lines & Asymptotes
| Equation of a line |
$\lim_{x→a}f(x)=f(a)=m·a+b$ |
| Horizontal asymptote |
The line $x=c$ is defined as $\lim_{x→c}f(x)=\pm ∞$ |
| Vertical asymptote |
The line $y=c$ is defined as $\lim_{x→\pm ∞}f(x)=c$ |
5.1.5 Squeeze Theorem
Given the transitive property, and three functions related as $f(x)≤g(x)≤h(x)$ converging to a point
$$\text{If }\lim_{x→a}f(x)=\lim_{x→a}h(x)=L\text{ then }\lim_{x→a}g(x)=L$$
5.1.6 Limits of Trig Functions 🔧
$$\lim\limits_{x→0}\frac{\sin(x)}{x}=1\qquad\lim\limits_{x→0}\frac{\cos(x)-1}{x}=0$$
Proof of Sine Limit
Using the incircle of a regular octagon with an apothem of magnitude 1 to define points in the sector for $0 \le \theta \le \pi/4$
Add to the top right corner a triangular region, forming an overall right triangle
The following determination can be made inside of triangle $OAD$
$$\text{arc} AC \lt |AB|+|BC| \lt \big(|AB|+|BD|=|AD|\big)$$
$$\land \medspace |AD|=|OA|·\tan(\theta)=\tan(\theta)$$
$$\therefore \text{arc} AC \lt\tan(\theta)$$
The apothem is 1 and the arc length $AC$ is the angle $\theta$. Substitute tangent for its cofunctions, then isolate $\cos(\theta)$
$$\theta \lt \frac{\sin(\theta)}{\cos(\theta)} \medspace → \medspace \cos(\theta) \lt \frac{\sin(\theta)}{\theta}$$
Connect point $C$ perpendicularly to $|OA|$ at point $E$, and connect a line segment to $|AC|$
Note that $|CE|=|OC|·\sin(\theta)=\sin(\theta)$ and $|CE|\lt |AC|\lt arc AC$. It follows that
$$\sin(\theta)\lt\theta\medspace\therefore\medspace\frac{\sin(\theta)}{\theta}\lt 1$$
Combine the two inequalities involving $\sin(\theta)/\theta$
$$\cos(\theta)\lt\frac{\sin(\theta)}{\theta}\lt 1$$
Apply the squeeze theorem for the limit on the right
$$\lim_{\theta→0^+}\cos(\theta)=1\medspace\land\medspace\lim_{\theta→0^+}1=1\medspace\therefore\medspace\lim_{\theta→0^+}\frac{\sin(\theta)}{\theta}=1$$
Sine is an odd function, so its nature tending to zero may be determined with sign reversal
$$\frac{\sin(-\theta)}{-\theta}=\frac{-\sin(\theta)}{-\theta}=\frac{\sin(\theta)}{\theta}$$
Since the function is the same from the reverse direction, the limit exists
$$\lim_{\theta→0^-}\frac{\sin(\theta)}{\theta}=\lim_{\theta→0^+}\frac{\sin(\theta)}{\theta}=\lim_{\theta→0}\frac{\sin(\theta)}{\theta}=1$$
Proof of Cosine Limit
5.1.7 Limits Related to Zero & Infinity
$$\lim\limits_{x→0^\pm} \frac{c}{x}=\pm ∞,\medspace c\ne 0$$
$$\lim\limits_{x→\pm ∞} \frac{c}{x}=0,\medspace c\ne 0$$
