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

Question

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?

Answer 1

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

Answer 2

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.