Vertical Asymptote (Definition §1.5.6)

The vertical line $x = a$ is called the vertical asymptote of the curve $y = f(x)$ if at least one of the following statements is true:

$$ \begin{equation} \begin{split} \lim_{x \to a^{-}} f(x) &= \infty \\ \lim_{x \to a^{-}} f(x) &= -\infty \\ \lim_{x \to a^{+}} f(x) &= \infty \\ \lim_{x \to a^{+}} f(x) &= -\infty \end{split} \end{equation} $$

Also:

$$\begin{equation} \begin{split} \lim_{x \to a} f(x) &= \lim_{x \to a^{-}} f(x) = \lim_{x \to a^{+}} f(x) = \infty \\ \lim_{x \to a} f(x) &= \lim_{x \to a^{-}} f(x) = \lim_{x \to a^{+}} f(x) = -\infty \end{split} \end{equation}$$

To my understanding, if there is a vertical asymptote at $x = a$, then $\lim_{x \to a}f(x)$ is technically undefined. But instead, of saying undefined, we say $\pm \infty$ since it gives more information. I.e. this is a special case of being undefined.

§1.6 | Calculating Limits Using the Limit Laws

Limit Laws

Suppose that $c$ is a constant and that the limits $\lim_{x \to a} f(x)$ and $\lim_{x \to a} g(x)$ exists, then:

$$ \begin{equation} \begin{split} \lim_{x \to a} \Big[ f(x) + g(x) \Big] &= \lim_{x \to a} f(x) + \lim_{x \to a} g(x) \\ \lim_{x \to a} \Big[ f(x) - g(x) \Big] &= \lim_{x \to a} f(x) - \lim_{x \to a} g(x) \\ \lim_{x \to a} \Big[ c \cdot f(x) \Big] &= c \cdot \lim_{x \to a} f(x) \\ \lim_{x \to a} \Big[ f(x) \cdot g(x) \Big] &= \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x) \\ \lim_{x \to a} \frac{f(x)}{g(x)} &= \frac{\lim_{x \to a} f(x)}{\lim_{x \to a}g(x)} \; \text{if} \lim_{x \to a}g(x) \neq 0 \end{split} \end{equation} $$

Power Law

$$\lim_{x \to a} \Big[ f(x) \Big]^n = \Big[ \lim_{x \to a} f(x) \Big]^n$$

$n$ must be a positive integer.

Root Law

$$\lim_{x \to a} \sqrt[n]{f(x)} = \sqrt[n]{\lim_{x \to a} f(x)}$$

$n$ must be a positive integer.

If $n$ is even, we assume that $\lim_{x \to a} f(x) > 0$.

§1.8 | Continuity

The Intermediate Value Theorem (IVT)

Suppose that $f$ is continuous on the closed interval $[a, b]$, and let N be any number between $f(a)$ and $f(b)$, where $f(a) \neq f(b)$, then there exists a number $c$ in $(a, b)$ such that $f(c) = N$.

Or alternatively, suppose $f$ is a function that is continuous at every point in the interval $[a, b]:$

$f$ will take on every value between $f(a)$ and $f(b)$ over the interval.
For any L between the values $f(a)$ and $f(b)$, there exists a number $c$ in $[a, b]$ for which $f(c) = L$.

The intermediate value theorem states that if a continuous function attains two values, it must also attain all values in between these two values.
The Intermediate Value Theorem states that a continuous function takes on every intermediate value between the function values $f(a)$ and $f(b)$.
The statement of the theorem has multiple requirements, all of which are necessary for the conclusion to hold.

The IVT can be used to show that a root exists, if given some interval between $a$ and $b$:

$f(a)$ is negative
$f(b)$ is positive

$f(a)$ is positive
$f(b)$ is negative

then therefore, if the function is continuous, then there must be a value for which some $x$ in $f(x) = 0$ exists. I.e. that a root exists. Which should make sense.

**More generally**, this can be extended to any value between some interval. For instance, if:

$f(a) = 3$

$f(a) = 9$

Then there must exist some value that we will call $x$, for which:

$f(x) = 6$

Chapter 1