Prove that $ f(x) = \tan x $ has infinitely many vertical asymptotes.

Question

Just by looking at $f(x)=\tan(x)$ graph one can tell it has infinite number of vertical asymptotes. However, I need to prove it in math-fashioned style, preferably with using the keyword $\lim$.

So far I came up with this:

$\tan\beta = \dfrac{\sin\beta }{\cos\beta } \implies \tan\beta $ is undefined when $\cos\beta = 0$

$\cos\beta=0$

$\beta = \frac{\pi}{2}+ k\pi$ and $k\in\mathbb{Z}$

Hence, vertical asymptotes of $f(x)$ are:

$x_{k} = \frac{\pi}{2} + k\pi$

For example: $x_{0}=\frac{\pi}{2},x_{5}=\frac{11\pi}{2},x_{-1}=\frac{-\pi}{2}, ...$

Is it a good enough proof solution? Can someone do better? How can I plot somewhere in here the $\lim$ "keyword"?

Thanks.

Answer 1

The idea is correct but for a formal proof we need to show that for any $k\in \mathbb{Z}$

$$\lim_{x\to \left(\frac{\pi}2+k\pi\right)^-} \frac{\sin x}{\cos x}=+\infty$$

$$\lim_{x\to \left(\frac{\pi}2+k\pi\right)^+} \frac{\sin x}{\cos x}=-\infty$$

To proceed for any k we can use $y=x-\left(\frac{\pi}2+k\pi\right) \to 0 \implies x=y+\left(\frac{\pi}2+k\pi\right)$ then use that

• $\sin x= \sin \left(\frac{\pi}2+k\pi+y\right)=\cos (k\pi+y)=\pm\cos y$
• $\cos x= \cos \left(\frac{\pi}2+k\pi+y\right)=-\sin (k\pi+y)=\mp \sin y$

and the limits become

$$\lim_{x\to \left(\frac{\pi}2+k\pi\right)^-} \frac{\sin x}{\cos x}=\lim_{y\to 0^-} -\frac{\cos y}{\sin y}$$

$$\lim_{x\to \left(\frac{\pi}2+k\pi\right)^-} \frac{\sin x}{\cos x}=\lim_{y\to 0^+} -\frac{\cos y}{\sin y}$$