# How do you locate the discontinuities of the function y = ln(tan(x)2)?

Apr 10, 2015

Assuming this says $y = \ln \left(2 \tan x\right)$:

Recall that $\ln x$ is continuous within its own domain (this is a tricky definition, but all it means is that its own domain happens to extend towards 0 without getting to 0, making it "continuous" because it never gets to a well-defined boundary). Its domain is $\left(0 , \infty\right)$, while its range is $\left(- \infty , \infty\right)$.

Also recall that $\tan x$ has a vertical asymptote at $x = \pm \frac{n \pi}{2}$ (it looks almost like $y = {x}^{3}$ with sharp vertical breaks), and is ZERO every integer $\boldsymbol{\pi}$. However, again, you could say it is continuous within its own domain because it has double-sided limits but does not ever truly reach a well-defined boundary.

The first thing to look at is where the ln function has discontinuities. The natural log function looks at the value in parentheses and evaluates the power to which e, Euler's number, must be raised to be equal to the value in the parentheses. For example:

$\ln \left(1\right) = 0$ This implies that ${e}^{0} = 1$ which makes sense. You can also see this equivalence by making both sides exponents of e.

From this example, it becomes clear that we will have a problem if we take the natural log of 0. What power could e be raised to to make it 0? The answer is a discontinuity, in this case the discontinuity is negative infinity. Since there is a discontinuity in the ln function at $\ln \left(0\right)$, finding any points where the interior function equals 0 will be finding discontinuities. To find the rest, find where the interior function has discontinuities.

As a result, this has discontinuities if

• $x = \pm \frac{n \pi}{2}$ --- this covers the case of $\pm n \pi$ as well as $\pm \frac{n \pi}{2}$, as we want $\pm \frac{\pi}{2} , \pm \pi , \pm \frac{3 \pi}{2}$, etc.

$n \in \mathbb{Z}$.