# How do you find discontinuity algebraically?

May 18, 2015

There is no universal method that works for all possible functions.

The problems beginning calculus students are presented usually involve either:

Rational functions and trigonomeric functions are continuous on their domain. (dincontinuous when we try to divide by $0$).

$f \left(x\right) = \frac{\left(x - 3\right) \left(x + 4\right)}{\left(x + 4\right) \left(x + 9\right)}$ is dincontinuous at $- 4$ and at $- 9$

Piecewise defined functions may be discontinuous where the rule changes. Check to be sure the limits from left and right are equal. And that these limits are equal to the value of the function at the $x$ value where the rule changes.
Examke 1
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$f \left(x\right) = \left\{\begin{matrix}{x}^{2} - 9 & \text{ & if " & x<4 \\ 2x - 1 & " & if } & x > 4\end{matrix}\right.$

$f$ is continuous to the left of $4$ (it is a polynomial -- they are continuous everywhere).
$f$ is continuous to the right of $4$ (it is a polynomial -- they are continuous everywhere)

Example 2
${\lim}_{x \rightarrow {4}^{-}} f \left(x\right) = {\left(4\right)}^{2} - 9 = 7$

${\lim}_{x \rightarrow {4}^{+}} f \left(x\right) = 2 \left(4\right) - 1 = 7$

Because the left and right limits are equa, we have:

${\lim}_{x \rightarrow 4} f \left(x\right) = 7$

But the function is not defined for $x = 4$ ($f \left(4\right)$ does not exist). so the function is not continuous at $4$. $f$ is defined and continuous "near' 4, so it is discontinuous at $4$.

Example 3
$g \left(x\right) = \left\{\begin{matrix}{x}^{2} - 9 & \text{ & if " & x<= 4 \\ 2x - 1 & " & if } & x > 4\end{matrix}\right.$ is continuous at $4$

Example 4
$g \left(x\right) = \left\{\begin{matrix}{x}^{2} - 9 & \text{ & if " & x<= 4 \\ 2x +6 & " & if } & x > 4\end{matrix}\right.$ is discontinuous at $4$.