Function Limits

All mathematical functions (as far I know) return $x \in [-\infty, \infty]$. However, some functions can be discontinual and have limits in the form of asymptotes. Inversely proportional functions are a example of this where all returns are forced between two asymptotes (often horizontal and vertical).

Horizontal, Vertical and Weird Asymptotes

In the case of a vertical asymptote, we will be able to discover a x value in which the y value approaches infinity. The oposite is true of a horizontal asymptote. The same is also true for asymptotes that are not strictly of 90 degree oposition, but the math of determining the limits is a bit more involved. There can also be more than two asymptotes in a function. It gets worse once you start adding functions together.

Continuity and Discontinuity

It is easy to think intuitively that any function $f(x)$ returns a value for any x, but this is not necessarily the case. Expressions that involve fractions can potentially mess up the function for certain x if it leads to zero in the divisor.

Example

The following function works as it is for every $x \in R \setminus {2}$:

$f(x) = \frac{3x^2 - 12}{x - 2}$

The reason for this is that we divide by zero if x = 2. But as this (as far we know) is only a problem if x = 2, we can try to work around it by aproaching 2 and see what values we get. Often we can also calculate f(2) accurately by rewriting the expression a different way.

Solutions and Notation

When we want to express aproaching a limit (2 in this case), we can write it the following way:

$f(x) = \lim\limits_{x \to 2} \frac{3x^2 - 12}{x - 2}$

Our task now is to calculate f(2) without dividing by zero. The first and simplest solution is often to factorize:

$f(x) = \lim\limits_{x \to 2} \frac{3x^2 - 12}{x - 2} = \frac{3(x - 2)(x + 2)}{x - 2} = 3(x + 2) = 3x + 6$ $f(2) = 3(2) + 6 = 12$

How many discontinuities?

There have to be as many discontinuities as there are zero points in the divisor’s expression. As our example’s divisor was a linear function, it only has one zero point, which mean we don’t have to worry about more discontinuities.

Some “lim” Rules

Premise: $\lim\limits_{x \to a} f(x) \land \lim\limits_{x \to a} g(x)$

  1. $\lim\limits_{x \to a}(f(x) + g(x)) = \lim\limits_{x \to a} f(x) + \lim\limits_{x \to a} g(x)$
  2. $\lim\limits_{x \to a}(f(x) * g(x)) = \lim\limits_{x \to a} f(x) * \lim\limits_{x \to a} g(x)$
  3. $\lim\limits_{x \to a} k * f(x) = k * \lim\limits_{x \to a} f(x)$
  4. $\lim\limits_{x \to a}\frac{f(x)}{g(x)} = \frac{\lim\limits_{x \to a} f(x)}{\lim\limits_{x \to a} g(x)} \land \lim\limits_{x \to a} g(x) \neq 0$ # Division by zero bad