Continuity is a property of a function. When you talk about continuity, you describe whether the graph of a function exists for all values of $x$ in an interval, and that these points are adjacent to each other, meaning there are no gaps between any of them.
When a graph is continuous, it means that you can draw it without lifting your pencil. By contrast, there are graphs where the $y$-value of a given $x$-value is some distance away from the $y$-value of an adjacent $x$-value. These graphs are not continuous. Functions that are not defined for all $x$-values on an interval are also not continuous on that interval. Below, you see drawings of the different cases:
Theory
If $\underset{x\to a}{\mathrm{lim}}f(x)=f(a)$, then $f$ is continuous for $x=a$.
If $\underset{x\to a}{\mathrm{lim}}f(x)\ne f(a)$, then $f$ is discontinuous for $x=a$.
When $f(x)$ is continuous for all $x$ in an interval, we say that it is continuous on the interval.
Example 1
Determine whether the function $f(x)=\frac{{x}^{2}-2x}{x-1}$ is continuous for $x=1$
$$\begin{array}{llll}\hfill \underset{x\to 1}{\mathrm{lim}}\frac{{x}^{2}-2x}{x-1}& =\underset{x\to 1}{\mathrm{lim}}\genfrac{}{}{1.0pt}{}{\phantom{\rule{0.17em}{0ex}}\frac{{x}^{2}}{{x}^{2}}-\frac{2x}{{x}^{2}}\phantom{\rule{0.17em}{0ex}}}{\phantom{\rule{0.17em}{0ex}}\frac{x}{{x}^{2}}-\frac{1}{{x}^{2}}\phantom{\rule{0.17em}{0ex}}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\underset{x\to 1}{\mathrm{lim}}\genfrac{}{}{1.0pt}{}{\phantom{\rule{0.17em}{0ex}}1-\frac{2}{x}\phantom{\rule{0.17em}{0ex}}}{\phantom{\rule{0.17em}{0ex}}\frac{1}{x}-\frac{1}{{x}^{2}}\phantom{\rule{0.17em}{0ex}}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\underset{x\to 1}{\mathrm{lim}}\frac{1-\frac{2}{1}}{\frac{1}{1}-\frac{1}{{1}^{2}}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{1-2}{1-1}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\text{notdefined}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$ You see that $f(x)$ is discontinuous, since you got an invalid expression when calculating the limit.
Example 2
Determine if the function $f(x)=\frac{{x}^{2}-2x+2}{x+3}$ is continuous for $x=-2$
$$\begin{array}{cc}\underset{x\to -2}{\mathrm{lim}}\frac{{x}^{2}-2x+2}{x+3}& \\ =\frac{{(-2)}^{2}-2(-2)+2}{-2+3}& \\ =\frac{4+4+2}{1}& \\ =10& \end{array}$$
You see that $f(x)$ is continuous since you got a numeric value as the result when you calculated the limit.