finite mathematics & applied calculus topic summary:

Introduction to the derivative: Part 2 of 3: Continuity

Introduction to the derivative: Part 2 of 3: Continuity

- If $a$ is an endpoint of the domain of $f$, then $\lim_{x \to a} f(x)$ is, as usual, understood to be the appropriate one-sided limit.
- If $a$ is an isolated point in the domain of $f$; that is, $f$ is defined at $a$ but at no other points within some distance of $a$ (for example, $f(x)=\sqrt{-x^2}$ is defined only at $x = 0$) then, even though there can be no limit at $a,$ we regard $f$ as continuous at $a.$

Checking continuity at $\bold{x = 1:}$ (red circle)
• $\displaystyle \lim_{x \to 1} f(x)$ exists, and equals $0.5$. ✓
Therefore, $f$ is continuous at $x = 1.$
• $f(1) = 0.5$ as well ✓ Checking continuity at $\bold{x = 0:}$ (blue circle)
• $\displaystyle \lim_{x \to 0} f(x)$ does not exist. ✗
Therefore, $f$ is not continuous at $x = 0.$
Checking continuity at $\bold{x = -1:}$ (green circle)
• $\displaystyle \lim_{x \to -1} f(x)$ exists, and equals $1$. ✓
Therefore, $f$ is not continuous at $x = -1.$
• $f(-1) = 0,$ which does not equal the limit. ✗ |

The function $f(x) = 3x^2-4x+2$ is a closed form function, and therefore is continuous on its domain (the set of all real numbers).

The function $\displaystyle g(x) = \frac{4x^2+1}{x-3}$ is also a closed form function, and therefore is continuous on its domain (the set of all real numbers except 3).The point $x = 3$ is not a discontinuity (as $x = 3$ is not in its domain) but is called a

Singularity at $\bold{a}$ | Discontinuity at $\bold{a}$ | |

The function $f(x) = \begin{cases} \dfrac{1}{x} & \text{if } x \ne 0 \\3 & \text{if } x = 0 \end{cases} \quad$ is defined at $x = 0$ and so is not singular there. It is, however, discontinuous at $x=0$.

Removable singularity at $\bold{a}$ | Essential singularity at $\bold{a}$ | |

$\displaystyle \lim_{x \to a}f(x) = 1$ | $\displaystyle \lim_{x \to a}g(x)$ dos not exist. | |

$f$ can be made continuous by defining $f(a) = 1$. | $g$ cannot be made continuous. | |

$\displaystyle f(x) = \frac{x^2-9}{x-3}$ has a singular point at $x = 3$. In Part 1 of this summary we saw that

- $\displaystyle \lim_{x \to 3} f(x) = \lim_{x \to 3} \frac{x^2-9}{x-3} = 6.$

- $\displaystyle \frac{x^2-9}{x-3} = \frac{(x-3)(x+3)}{x-3} = x+3$,