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 $\cos\beta = 0$


$\beta = \frac{\pi}{2}+ k\pi$ 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"?


  • 2
    $\begingroup$ No, unfortunately it's not good enough. A vertical asymptote doesn't mean the same thing as "undefined." Try to write down what a vertical asymptote means. Then you'll see how the "\lim" comes in. $\endgroup$ – saulspatz Nov 24 '18 at 20:49
  • 1
    $\begingroup$ Well, the graph of $\tan x$ is drawn using the information that it has infinitely many vertical asymptotes at $x=n\pi/2$ where $n$ is odd and not the other way round. $\endgroup$ – Paramanand Singh Nov 25 '18 at 1:02

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

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

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

  • $\begingroup$ Thanks! How can I further solve it? Or is that already completed solution just like that? $\endgroup$ – weno Nov 24 '18 at 21:01
  • $\begingroup$ @weno 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 $$\sin x= \sin \left(\frac{\pi}2+k\pi+y\right)=\cos (k\pi+y)=\pm\cos y$$ and $$\cos x= \cos \left(\frac{\pi}2+k\pi+y\right)=-\sin (k\pi+y)=\mp \sin y$$ I add that to the answer. $\endgroup$ – user Nov 24 '18 at 21:13
  • $\begingroup$ Thank you for your effort. If that's not too much to ask: are these '' - '' signs correct in the last line? Here I marked it with red circles: imgur link Also is that a completed solution, or do I need to solve the last line? Thanks again, and have a good day! $\endgroup$ – weno Nov 24 '18 at 21:33
  • $\begingroup$ @weno Well when $x\to \left(\frac{\pi}2+k\pi\right)^-$ we have $y=x-\left(\frac{\pi}2+k\pi\right) \to 0^-$ and similarly for the other side. Yes also the second sign is correct indeed for $k$ even we have $\cos (k\pi+y)=\cos y$ but $-\sin (k\pi+y)=-\sin y$ and for $k$ odd both sign reverse. Yes I think that that can be considered a complete proof given that the asymptotes are only at those points. $\endgroup$ – user Nov 24 '18 at 21:40

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