# Difference between revisions of "Math 22 Continuity"

## Continuity

Informally, a function is continuous at ${\displaystyle x=c}$ means that there is no interruption in the graph of ${\displaystyle f}$ at ${\displaystyle c}$.

## Definition of Continuity

 Let ${\displaystyle c}$ be a real number in the interval ${\displaystyle (a,b)}$, and let ${\displaystyle f}$ be a function whose domain contains the interval ${\displaystyle (a,b)}$ . The function ${\displaystyle f}$ is continuous at ${\displaystyle c}$ when
these conditions are true.
1. ${\displaystyle f(c)}$ is defined.
2. ${\displaystyle \lim _{x\to c}f(x)}$ exists.
3. ${\displaystyle \lim _{x\to c}f(x)=f(c)}$
If ${\displaystyle f}$ is continuous at every point in the interval ${\displaystyle (a,b)}$, then ${\displaystyle f}$ is continuous on the open interval ${\displaystyle (a,b)}$.


## Continuity of piece-wise functions

Discuss the continuity of ${\displaystyle f(x)={\begin{cases}x+2&{\text{if }}-1\leq x<3\\14-x^{2}&{\text{if }}3\leq x\leq 5\end{cases}}}$

Solution:
On the interval ${\displaystyle [-1,3)}$, ${\displaystyle f(x)=x+2}$ and it is a polynomial function so it is continuous on ${\displaystyle [-1,3)}$
On the interval ${\displaystyle [3,5]}$, ${\displaystyle f(x)=14-x^{2}}$ and it is a polynomial function so it is continuous on ${\displaystyle [3,5]}$
Finally we need to check if ${\displaystyle f(x)}$ is continuous at ${\displaystyle x=3}$.
So, consider ${\displaystyle \lim _{x\to 3^{-}}f(x)=\lim _{x\to 3^{-}}x+2=3+2=5}$
Then, ${\displaystyle \lim _{x\to 3^{+}}f(x)=\lim _{x\to 3^{+}}14-x^{2}=14-(3)^{2}=5}$.
Since ${\displaystyle \lim _{x\to 3^{-}}f(x)=5=\lim _{x\to 3^{+}}f(x)}$, \lim_{x\to 3} f(x) exists.
Also notice ${\displaystyle f(3)=14-(3)^{2}=5=\lim _{x\to 3}f(x)}$
So by definition of continuity, ${\displaystyle f(x)}$ is continuous at ${\displaystyle x=3}$.
Hence, ${\displaystyle f(x)}$ is continuous on ${\displaystyle [-1,5]}$

## Types of Discontinuity

Removable discontinuity: If the function factors and the bottom term cancels, the discontinuity at the x-value for which the denominator was zero is removable, so the graph has a hole in it. For example: ${\displaystyle f(x)={\frac {x^{2}-2x-3}{x+1}}={\frac {(x-3)(x+1)}{x+1}}=x-3}$. This function ${\displaystyle f(x)}$ is y=x-3 with a hole at ${\displaystyle x=-1}$ since ${\displaystyle x=-1}$ makes ${\displaystyle f(x)}$ undefined.

Infinite discontinuity: An infinite discontinuity exists when one of the one-sided limits of the function is infinite and the limit does not exist. This is an infinite discontinuity. In another word, we have infinite discontinuity when either ${\displaystyle \lim _{x\to a^{-}}f(x)=\pm \infty }$ or ${\displaystyle \lim _{x\to a^{+}}f(x)=\pm \infty }$

Jump discontinuity: The function is approaching different values depending on the direction ${\displaystyle x}$ is coming from. When this happens, we say the function has a jump discontinuity at ${\displaystyle x=a}$. In another word, ${\displaystyle \lim _{x\to a^{-}}f(x)\neq \lim _{x\to a^{+}}f(x)}$

## Notes

Polynomial function is continuous on the entire real number line (ex: ${\displaystyle f(x)=2x^{2}-1}$ is continuous on ${\displaystyle (-\infty ,\infty )}$)

Rational functions is continuous at every number in its domain. (ex: ${\displaystyle f(x)={\frac {x+2}{x^{2}-1}}}$ is continuous on ${\displaystyle (-\infty ,-1)\cup (-1,1)\cup (1,\infty )}$ since the denominator cannot equal to zero)