2
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.