
The limit of a continuous function at a point is equal to the value of the function at that point.

Limits are simple to compute when they can be found by plugging the value into the function. That is, when We call this property continuity.
Consider the graph of the function $f$ Which of the following are true?
$f$ is continuous at $x=0.5$ $f$ is continuous at $x=1$ $f$ is continuous at $x=1.5$

It is very important to note that saying

“a function $f$ is continuous at a point $a$

is really making three statements:

(a)
$f(a)$ is defined. That is, $a$ is in the domain of $f$.
(b)
$\lim _{x\to a} f(x)$ exists.
(c)
$\lim _{x\to a} f(x) = f(a)$.

The first two of these statements are implied by the third statement.

Building from the definition of continuity at a point, we can now define what it means for a function to be continuous on an open interval.

Loosely speaking, a function is continuous on an interval $I$ if you can draw the function on that interval without any breaks in the graph. This is often referred to as being able to draw the graph “without picking up your pencil.”

Compute: $\lim _{x\to \pi } x^3\begin {prompt}= \answer {\pi ^3}\end {prompt}$
Compute: $\lim _{x\to \pi } \sqrt {2}\begin {prompt}= \answer {\sqrt {2}}\end {prompt}$
Compute: $\lim _{x\to \pi } \cos {x}\begin {prompt}= \answer {-1}\end {prompt}$

### Left and right continuity

At this point we have a small problem. For functions such as $\sqrt {x}$, the natural domain is $0\leq x <\infty$. This is not an open interval. What does it mean to say that $\sqrt {x}$ is continuous at $0$ when $\sqrt {x}$ is not defined for $x<0$? To get us out of this quagmire, we need a new definition:

This allows us to talk about continuity on closed and half-closed intervals.

Here we give the graph of a function defined on $[0,10]$. Select all intervals for which the following statement is true.

The function $f$ is continuous on the interval $I$.

$I=[0,10]$ $I=[0,4]$ $I=[4,6]$ $I=[4,6)$ $I= (4,6]$ $I= (4,6)$ $I=(6,10]$ $I=[6,10)$