# Rigorously proving $\lim_{x \to (2n+1)^+} \tan\left(\frac{\pi x} 2\right) = - \infty$

How to rigorously prove this limit? $$\lim_{x \to (2n+1)^+} \tan\left(\frac{\pi x} 2\right) = - \infty$$

My studying book used function composition and Heine definition for limits (Using sequences) to do so. But I didn't really understand the proof.

How can I prove this in a rigorous way?

## 2 Answers

Since:

• $$\displaystyle\tan\left(\frac{\pi x}2\right)=\frac{\sin\left(\frac{\pi x}2\right)}{\cos\left(\frac{\pi x}2\right)}$$;
• $$\displaystyle\lim_{x\to(2n+1)^+}\sin\left(\frac{\pi x}2\right)=\begin{cases}1&\text{ if n is even}\\-1&\text{ if n is odd;}\end{cases}$$
• if $$x$$ is close to and greater than $$2n+1$$, then $$\displaystyle\cos\left(\frac{\pi x}2\right)\begin{cases}>0&\text{ if n is even}\\<0&\text{ if n is odd}\end{cases}$$

you have$$\lim_{x\to(2n+1)^+}\tan\left(\frac{\pi x}2\right)=-\infty.$$

As $$f(x)= \tan\left(\frac{\pi x} 2\right)$$ is a periodic function of period $$2$$, the limit is equal to

$$\lim_{x \to 1^+} \tan\left(\frac{\pi x} 2\right).$$

And

$$\lim_{x \to 1^+} \sin\left(\frac{\pi x} 2\right)=1$$ while $$\lim_{x \to 1^+} \cos\left(\frac{\pi x} 2\right)=0$$ by taking only negative values in the interval $$(1,2)$$.

This provides the expected limit.