A function f(x) is **continuous** at a point x = a if and only if the following three conditions are satisfied:

- i. It must be defined at that point, i.e., f(a) is defined.
- ii. Its limit must exist at the point, $\lim_{x \to a} f(x)$ exists. The left and right-hand limits are defined and equal.
- iii. The value of the function at that point must equal the value of the limit at that point, $\lim_{x \to a} f(x) = f(a).$

Let f(x) be a function defined on an interval that contains x = a, then we say that f(x) is continuos at x = a, if

$\forall \epsilon>0, \exists \delta>0: |f(x)-f(a)|<\epsilon, whenever~ |x-a| < \delta.$

A function is **discontinuous if we cannot draw its graph without lifting your pen**. The function’s graph does not form a continuos line, but instead has a break or gap. More formally, a function is discontinuous at a point a if it fails to be continuous at at, that is, when it breaks any of the continuity criteria, such as, f(a) is not defined, $\lim_{x \to a} f(x)$ does not exist, or $\lim_{x \to a} f(x) ≠ f(a).$

There are different types of discontinuities:

- A
**jump discontinuity**is a discontinuity where the value of a function ‘jumps’ from one value to another, that is,**where the right hand limit and left hand limit both exist, but these limits are not equal**to each other. Examples:

$f(x) = \begin{cases} x + 1, &x > 0\\\\ -2x + 2, &x < 0 \end{cases}$ -1.g.-

$\lim_{x \to 0^{+}} f(x) = 1 ≠ 2 = \lim_{x \to 0^{-}} f(x).$

$f(x) = \begin{cases} -x^{2} + 4, &x≤3\\\\ 4x - 8, &x > 3 \end{cases}$

$\lim_{x \to 3^{+}} f(x) = 4 ≠ -5 = \lim_{x \to 3^{-}} f(x).$

- A
**removable discontinuity**is a discontinuity that results when**the limit of a function exists**but is not equal to the value of the function at the given point. It seems that the function is begging to be redefined. f(x)=^{sin(x)}⁄_{x}is a good example -1.h-. $\lim_{x \to 0^{+}} \frac{sin(x)}{x} = 0$, but f(0) is not defined.

f(x) = $\frac{x^{2}-4}{x-2}$. $\lim_{x \to 2}\frac{x^{2}-4}{x-2} = \lim_{x \to 2}\frac{(x-2)(x+2)}{x-2} = \lim_{x \to 2} (x+2) = 4,$ but f(2) is undefined.

$f(x) = \begin{cases} 2x + 1, &x<1\\\ 2, &x=1\\\ -x + 4, &x > 1 \end{cases}$

$\lim_{x \to 1^{+}} f(x) = 3 = \lim_{x \to 1^{-}} f(x), but \lim_{x \to 1} f(x) = 3 ≠ 2 = f(1)$

- f has an
**infinite discontinuity**at a if $\lim_{x \to a^{+}} f(x) = \pm\infty$ or $\lim_{x \to a^{-}} f(x) = \pm\infty.$ Examples:

y = ^{1}⁄_{x} -1.f.- where $\lim_{x \to 0^{+}} \frac{1}{x} = \infty$ and $\lim_{x \to 0^{-}} \frac{1}{x} = -\infty$

y = ^{(x+2)}⁄_{(x+1)} where $\lim_{x \to -1^{+}} \frac{x+2}{x+1} = \infty$ and $\lim_{x \to -1^{-}} \frac{x+2}{x+1} = -\infty$

- A mixed discontinuity is a discontinuity that results when
**at least one of the one-sided limits does not exist.**An example is y = sin(^{1}⁄_{x}) -1.i.-

Theorem. If a function f is differentiable at x = a, then f is continuous at x = a.

Proof. We need to check if $\lim_{x \to a} f(x) = f(a) ↔ \lim_{x \to a} f(x) - f(a) = 0$

$\lim_{x \to a} f(x) - f(a) = \lim_{x \to a} \frac {f(x) - f(a)}{x - a}(x-a) = f’(a)·0 = 0.$