aka Relations of Distance
Definition: We can make x get close to a by making $| x - 1 |$ sufficiently close to a (but not a)
$ \lim\limits_{x \to a} f(x) = L$
$ | f(x) - L | $
Note: usually f(a) is not defined when dealing with limits
Example: $ \lim\limits_{x \to 1} 2x +1 $
Conjecture: $ \lim\limits_{x \to 1} 2x +1 = 3 $
$| (2x + 1) - 3 | = | 2x -2 | = 2| x - 1 |$
$\to$ we can make $|(| 2x + 1) -3|$ very close to 0 by making | x - 1 | very close to 0.
Example: $ \lim\limits_{x \to 1} \frac{x^2 - 1}{x - 1}$
$\frac{x^2 - 1}{x - 1} = \frac{(x + 1)(x -1)}{( x - 1 )} = x + 1 $
Conjecture:
$ \lim\limits_{x \to 1} \frac{x^2 - 1}{x - 1} = 2 $
$| \frac{x^2 - 1}{x - 1} - 2 | = | \frac{x^2 - 1 -2(x - 1)}{x - 1} | =| \frac{x^2 - 2x + 1)}{x - 1} | = \frac{(x + 1)(x - 1)}{x - 1}| = | x + 1 | $
$\to$ we can make $ \lim\limits_{x \to 1} \frac{x^2 - 1}{x - 1} $ very close to 0 by making | x + 1 | very close to 0.
$ \begin{array}{rrrrrrrr} \text{Sum } & \lim\limits_{x \to a }(f(x) + g(x)) = \lim\limits_{x \to a } f(x) + \lim\limits_{x \to a } g(x) = L_1 + L_2 \\ \text{Sub } & \lim\limits_{x \to a } (f(x) - g(x)) = \lim\limits_{x \to a } f(x) - \lim\limits_{x \to a } g(x) = L_1 + L_2 \\ \text{Constant Multiplication } & c \in \mathbb{R}, \lim\limits_{x \to a } (c.f(x)) = c \times \lim\limits_{x \to a } f(x) = x \cdot L_1 \\ \text{Product } & \lim\limits_{x \to a } (f(x) \times g(x)) = \lim\limits_{x \to a } f(x) \cdot \lim\limits_{x \to a } g(x) L_1 \times L_2 \\ \text{Quotient } & \lim\limits_{x \to a } (f(x) / g(x)) = \lim\limits_{x \to a } f(x) / \lim\limits_{x \to a } g(x) = L_1 / L_2 \\\ \end{array} $
Suppose $ f(x) \leq g(x) $ for x close to a. then
$ \lim\limits_{x \to a } f(x) \leq \lim\limits_{x \to a } g(x) $
Suppose $f(x) \leq g(x) \leq h(x) $
if $\lim\limits_{x \to a } f(x) = \lim\limits_{x \to a } h(x)$
then $ \lim\limits_{x \to a } f(x) = \lim\limits_{x \to a } g(x) = \lim\limits_{x \to a } h(x) $