# Is f(x)=(x^2-9)/(x-3) continuous at x=3?

Dec 7, 2016

$f \left(x\right) = \frac{{x}^{2} - 9}{x - 3}$ is not continuous at $x = 3.$

#### Explanation:

In order for a function $f \left(x\right)$ to be continuous at a given $x$-value $a$, the following condition must be satisfied:

$\textcolor{g r a y}{\left[1\right]} {\lim}_{x \to a} f \left(x\right) = f \left(a\right)$

What this is saying is that, as $x$ gets closer to $a$, $f \left(x\right)$ should also get closer to $f \left(a\right)$.

For the given function $f \left(x\right)$, the limit on the left-hand side of $\textcolor{g r a y}{\left[1\right]}$ will evaluate correctly. You'll end up with ${\lim}_{x \to 3} \frac{{x}^{2} - 9}{x - 3} = 6.$ However, the right-hand side of $\textcolor{g r a y}{\left[1\right]}$ presents a problem: what is $f \left(x\right)$ when x=3? The answer is, it is not defined, because at that point, we have $f \left(x\right)$ "equal" to $\frac{0}{0}$:

$f \left(3\right) = \frac{{3}^{2} - 9}{3 - 3} = \frac{9 - 9}{0} = \frac{0}{0}$

And this "value" of $\frac{0}{0}$ is indeterminate.

Thus, the function "breaks" at $x = 3$, and so, because there is no $f \left(3\right)$, it is not possible to say ${\lim}_{x \to 3} f \left(x\right) = f \left(3\right) .$ Thus, $f \left(x\right)$ is not continuous at $x = 3.$

## Bonus:

Here's a tip to determine where some functions will be discontinuous. Any $x$-value that makes any denominator in the function equal to $0$ is a point of discontinuity. So for the function above, $x - 3 = 0$ when $x = 3$, and so the function will be discontinuous at $x = 3.$

The only exception to this is if the function is piecewise defined at potential "breaking" points, like this:

${f}^{\star} \left(x\right) = \left\{\begin{matrix}\frac{{x}^{2} - 9}{x - 3} & \text{ & " & x!=3 \\ 6 & " & } & x = 3\end{matrix}\right.$

In this case, since ${f}^{\star} \left(x\right)$ is defined at $x = 3$, we have ${\lim}_{x \to 3} {f}^{\star} \left(x\right) = {f}^{\star} \left(3\right) ,$ so ${f}^{\star} \left(x\right)$ is continuous at $x = 3$ (and everywhere).