What are the removable and non-removable discontinuities, if any, of f(x)=(x+3)/((x-4)(x+3))?

Aug 2, 2017

The discontinuities of this function are at $x = 4$ and $x = - 3$. The one at $x = - 3$ is removable.

Explanation:

The function $f \left(x\right) = \frac{x + 3}{\left(x - 4\right) \left(x + 3\right)}$ is undefined at $x = 4$ and $x = - 3$, so it has discontinuities at those two values of $x$.

The one at $x = - 3$, however, is "removable", because, after cancellation, $f \left(x\right) = \frac{1}{x - 4}$ whenever $x$ is not equal to $- 3$. The expression $\frac{1}{x - 4}$ is defined at $x = - 3$ and has a value of $\frac{1}{- 3 - 4} = - \frac{1}{7}$ there.

Because of this, ${\lim}_{x \to - 3} f \left(x\right) = - \frac{1}{7}$ and the graph of $f \left(x\right) = \frac{x + 3}{\left(x - 4\right) \left(x + 3\right)}$ only has a "hole" at $x = - 3$ (no vertical asymptote). This hole can be "filled in" by defining $f \left(- 3\right) = - \frac{1}{7}$ (in other words, make $f$ a piecewise-defined function).

By filling in this hole in the graph of $f$, we have "removed" the original discontinuity there. That's why it's called a removable discontinuity.

The discontinuity at $x = 4$ is not removable. The graph of $f$ has a vertical asymptote there. In fact, ${\lim}_{x \to 4 +} f \left(x\right) = + \infty$ ($x$ approaches 4 from the right) and ${\lim}_{x \to 4 -} f \left(x\right) = - \infty$ ($x$ approaches 4 from the left).

The graph of $f$ is shown below. The hole is not seen because the computer is just connecting the dots. The vertical asymptote can be seen, however.

graph{(x+3)/((x-4)(x+3)) [-10.13, 10.145, -5.07, 5.065]}