# How do I find one-sided limits of trig functions?

Jul 30, 2015

At points in the domain of a trigonometric function, it will be continuous, and you can evaluate one-sided limits just like two-sided limits: by substituting what $x$ is approaching into the function. Maybe you meant to ask about inverse trigonometric functions though? (See the explanation below)

#### Explanation:

A trigonometric function like $f \left(x\right) = \sin \left(x\right)$ is defined and continuous for all real numbers $x$. Therefore, if $a$ is a specific real number,

${\lim}_{x \to a +} f \left(x\right) = {\lim}_{x \to a -} f \left(x\right) = {\lim}_{x \to a} f \left(x\right) = f \left(a\right)$.

For example, if $a = \frac{\pi}{3}$, then

${\lim}_{x \to \frac{\pi}{3} +} f \left(x\right) = {\lim}_{x \to \frac{\pi}{3} -} f \left(x\right) = {\lim}_{x \to \frac{\pi}{3}} f \left(x\right) = \sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$,

If $f \left(x\right) = \csc \left(x\right) = \frac{1}{\sin \left(x\right)}$, then this is defined and continuous for all real numbers $x$ except for integer multiples of $\pi$ ($0 , \setminus \pm \pi , \setminus \pm 2 \pi , \setminus \pm 3 \pi , \setminus \ldots$). And, for example, if $a = \frac{\pi}{3}$:

${\lim}_{x \to \frac{\pi}{3} +} \csc \left(x\right) = {\lim}_{x \to \frac{\pi}{3} -} \csc \left(x\right) = {\lim}_{x \to \frac{\pi}{3}} \csc \left(x\right)$

$= \csc \left(\frac{\pi}{3}\right) = \frac{2}{\sqrt{3}} = \frac{2 \sqrt{3}}{3}$.

On the other hand, the inverse trigonometric function $f \left(x\right) = {\sin}^{- 1} \left(x\right) = \arcsin \left(x\right)$ has a domain equal to the closed interval $\left[- 1 , 1\right] = \left\{x : - 1 \setminus \le q x \setminus \le q 1\right\}$. In this case, we are only "allowed" to let $x$ approach $- 1$ from the right and $1$ from the left. In these cases, we still end up evaluating the limit by substitution (the graph is "one-sided continuous" in those cases):

${\lim}_{x \to - 1 +} {\sin}^{- 1} \left(x\right) = {\sin}^{- 1} \left(- 1\right) = - \frac{\pi}{2}$ and

${\lim}_{x \to 1 -} {\sin}^{- 1} \left(x\right) = {\sin}^{- 1} \left(1\right) = \frac{\pi}{2}$

Another common set of one-sided limits to know are those for ${\cos}^{- 1} \left(x\right) = \arccos \left(x\right)$:

${\lim}_{x \to - 1 +} {\cos}^{- 1} \left(x\right) = {\cos}^{- 1} \left(- 1\right) = \pi$ and

${\lim}_{x \to 1 -} {\cos}^{- 1} \left(x\right) = {\cos}^{- 1} \left(1\right) = 0$

For ${\tan}^{- 1} \left(x\right) = \arctan \left(x\right)$, you have to let $x$ approach $\setminus \pm \setminus \infty$ "from one-side" (though you don't bother putting this in the notation). Technically, we are not doing any "substituting" of $\setminus \infty$ here. We are just using knowledge about how ${\tan}^{- 1} \left(x\right)$ is defined and how the vertical asymptotes of $\tan \left(x\right)$ correspond to horizontal asymptotes of ${\tan}^{- 1} \left(x\right)$.

${\lim}_{x \to \setminus \infty} {\tan}^{- 1} \left(x\right) = \frac{\pi}{2}$ and ${\lim}_{x \to - \setminus \infty} {\tan}^{- 1} \left(x\right) = - \setminus \frac{\pi}{2}$.