# Continuity

Limits logically come before continuity since the definition of continuity requires using limits. But practically and historically, continuity comes first. The concept of a limit is used to explain the various kinds of discontinuities and asymptotes. Start by studying discontinuities.

Types of discontinuities to consider: removable (a gap or hole in the graph), jump, infinite (vertical asymptotes), oscillating, and end behavior (horizontal asymptotes).

Numerically: Make a table for the value of $\displaystyle \frac{1}{x-3}$ near x = 3 and as $\displaystyle x\to \pm \infty$. Relate the values and their signs to the graph. (Divide by a small number get a big number; divide by a big number, get a small number.)

Use the vocabulary of limits to explain the features of graphs. Example: The function $\displaystyle \frac{{{x}^{2}}-4}{x-2}$ has no value at x = 2 (f(2) does not exist), but as you get closer to x = 2 the function value gets closer to 4 ($\displaystyle \underset{x\to 2}{\mathop{\lim }}\,f\left( x \right)=4$).

Relate the limit, value and graph of the function. In the example above, the graph looks like the line $y=x+2$ with a gap or hole at the point (2, 4). Another example: $\displaystyle \underset{x\to \infty }{\mathop{\lim }}\,\frac{3{{x}^{2}}+x+8}{{{x}^{2}}+1}=3$ since, the graph gets closer to y = 3 as you go farther to the right. The line y = 3 is a horizontal asymptote.

Do this numerically as well:  $\displaystyle \frac{3{{x}^{2}}+x+8}{{{x}^{2}}+1}=3+\frac{x+5}{{{x}^{2}}+1}$ and since the fraction gets smaller as |x| gets larger, the function approaches 3 from above when x > 0 and from below when x < 0 (why?)

Extra for your experts: Discuss the reason for the jump discontinuity of

$\displaystyle f\left( x \right)=\frac{\cos \left( x \right)\sqrt{{{x}^{2}}-2x+1}}{x-1}$