Video Notes on Limits and Continuity

f_1(x) = \sqrt{x + 13}
- This function, f_1(x), calculates the square root of the value $x$ plus 13.
- Domain: x \ge -13
- Graph notes: starts at the point (-13, 0) and increases as x grows; for example, f_1(3) = \sqrt{3 + 13} = 4
- Shape: increasing, concave down (typical of a square-root function)
f_2(x) = \dfrac{1}{x + 2}
- This function, f_2(x), is a rational function, meaning it's a ratio of two polynomials. It calculates 1 divided by the sum of $x$ and 2.
- Graph notes: hyperbola with a vertical asymptote at x = -2; on the right side (x > -2) the branch is positive and descends toward 0 as x \to \infty; on the left side (x < -2) the branch is negative and ascends toward 0 as x \to -\infty

Consider f_2(x) = \dfrac{1}{x + 2}
Right-hand limit: \lim_{x \to -2^+} \dfrac{1}{x + 2} = +\infty
Left-hand limit: \lim_{x \to -2^-} \dfrac{1}{x + 2} = -\infty
General limit: \lim_{x \to -2} \dfrac{1}{x + 2} does not exist (infinite discontinuity)
Interpretation: vertical asymptote at x = -2; the function values blow up to opposite infinities on each side

The transcript mentions a point where an open circle would be, but the point is filled in by another value
Conceptual meaning: this is related to removable discontinuities
If a graph would have a hole at x = a (limit exists) but the function value f(a) is defined differently or not defined, the function is not continuous at a
If a filled dot at x = a matches the limit value, the function is continuous at a
If the filled dot does not match the limit, there is a removable discontinuity or a jump depending on the configuration

Sine and cosine are continuous everywhere
- Intuition from the transcript: you can draw sin and cos without lifting your pencil; that matches the formal idea that they are continuous for all real x
- Formal statement: for all x \in \mathbb{R}, \lim{h \to 0} \sin(x + h) = \sin x and \lim{h \to 0} \cos(x + h) = \cos x
- Therefore, sin x and cos x are continuous on \mathbb{R}
Polynomials are continuous everywhere
- Quote from the transcript: "polynomials were continuous everywhere"
- Formalization: if p(x) is a polynomial, then for every a, \lim_{x \to a} p(x) = p(a). This means that for any polynomial representation p(x), the limit of the function as x approaches a point a is simply the function's value at that point a.
Rational functions and domain constraints
- A rational function r(x) = \dfrac{p(x)}{q(x)} (a fraction where p(x) and q(x) are polynomials, and q(x) is not zero) is continuous at a point a provided q(a) \neq 0
- If q(a) = 0, the function is not continuous at a (the domain excludes a)
- The transcript notes: "this is a rational function. Clearly, it's not continuous everywhere. Negative two fails. This is not continuous everywhere because there's anything exist."—emphasizing the vertical asymptote at x = -2 where the denominator is zero

Infinite behavior near vertical asymptotes
- When a function tends to \pm \infty near a point, that indicates a vertical asymptote and an improper limit
- Example from the transcript: near x = -2 for f_2, the function heads toward \text{-}\infty from the left and +\infty from the right
- Consequence: the overall limit at that point does not exist

To determine limit existence at a point, compare one-sided limits:
- If both sides approach the same finite value L, then \lim_{x \to a} f(x) = L and the function is continuous at a (provided f(a) = L)
- If the one-sided limits diverge to opposite infinities or do not agree, the limit does not exist
Core definitions touched on in the transcript
- Continuity: a function is continuous at a point a if \lim_{x \to a} f(x) = f(a)
- Vertical asymptotes: points where the function diverges to \pm \infty as x approaches a given value
- Holes and removable discontinuities: a hole indicates a point not in the domain; a filled dot at the same x-value may or may not match the limit value

Square-root function: f_1(x) = \sqrt{x + 13} This function means the square root of $x$ plus 13. Its domain is x \ge -13, so $x$ must be greater than or equal to -13.
Example value: f_1(3) = \sqrt{3 + 13} = 4 This is an example of the square root of $3$ plus $13$, which equals $4$.
Rational function: f_2(x) = \dfrac{1}{x + 2} This function means $1$ divided by the quantity $x$ plus $2$.
Vertical asymptote at: x = -2 This marks the vertical asymptote where $x$ equals -2.
One-sided limits for the rational function: \lim{x \to -2^-} \dfrac{1}{x + 2} = -\infty (The left-hand limit as $x$ approaches -2 from values less than -2, results in negative infinity), \lim{x \to -2^+} \dfrac{1}{x + 2} = +\infty (The right-hand limit as $x$ approaches -2 from values greater than -2, results in positive infinity).
Limit at the vertical asymptote does not exist: \lim_{x \to -2} \dfrac{1}{x + 2} \text{ does not exist} (The general limit as $x$ approaches -2 does not exist because the one-sided limits approach different infinities).
Sine and cosine continuity statements: \text{For all } x \in \mathbb{R}, \quad \lim{h \to 0} \sin(x + h) = \sin x, \quad \lim{h \to 0} \cos(x + h) = \cos x (These statements show that for all real numbers $x$, the limit of $\sin(x + h)$ as $h$ approaches zero is $\sin x$, and similarly for $\cos(x + h)$ is $\cos x$, demonstrating their continuity).
Polynomials: continuous on \mathbb{R} (all real numbers); rational functions continuous where the denominator is nonzero.

Continuity is a foundational concept in calculus, underpinning derivatives and integrals
Understanding where functions are continuous helps in graphing, evaluating limits, and solving problems involving composition of functions
The idea of holes vs filled points connects to the concept of function definition on a domain and how the limit behavior relates to the actual function value at a point