# Proving $\lim_{x \to 1} \left (\tan \left(\frac{\pi x}{2} \right ) \lfloor x \rfloor \right )$ doesn't exist

I'm asked to prove that the following limit doesn't exist: $$\lim_{x \to 1} \left (\tan \left(\frac{\pi x}{2} \right ) \lfloor x \rfloor \right )$$

My attempt was to break the case into one-sided limits. So, for example, I could prove that

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

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

and then show that $$\lim_{x \to 1^+} \left (\tan \left(\frac{\pi x}{2} \right ) \lfloor x \rfloor \right ) \neq \lim_{x \to 1^-} \left (\tan \left(\frac{\pi x}{2} \right ) \lfloor x \rfloor \right )$$

using the product rule for infinite limits.

The only problem is that I'm not allowed to use the theorem about the limit of a composition of functions ($g(t)=\pi t /2$ and $f(x)=\tan{x}$) nor any theorem related to the limit of a continuous function when proving $\lim_{x \to 1^+} \tan \left(\frac{\pi x}{2} \right ) = -\infty$. I can however use the fact that $\lim_{x \to \frac{\pi}{2}^+}\tan x=-\infty$ and $\lim_{x \to \frac{\pi}{2}^-}\tan x=\infty$. Is there any way I can prove it using the basic limit arithemtic rules?

Hint. You may observe that, as $x \to 1$, $$\tan \left( \frac \pi2x\right)=-\frac{2}{\pi (x-1)}+\mathcal{O}(x-1) \tag1$$ and that $$\lfloor x \rfloor =\begin{cases} 1 & \text{if } x \to 1^+ \\ 0 & \text{if } x \to 1^- \end{cases}$$ giving $$\tan \left( \frac \pi2x\right)\lfloor x \rfloor \to\begin{cases} -\infty & \text{if } x \to 1^+ \\ 0 & \text{if } x \to 1^- \end{cases}$$ thus a limit doesn't exist as $x \to 1$.

To see $(1)$, we may write, as $x \to 1$, \begin{align} \tan \left( \frac \pi2x\right)&=\tan \left( \frac \pi2(x-1)+\frac \pi2\right)\\\\&=\frac{\cos\frac \pi2(x-1)}{-\sin\frac \pi2(x-1)}\\\\ &=\frac{1+\mathcal{O}((x-1)^2)}{-\sin\frac \pi2(x-1)}\\\\ &=\frac{1+\mathcal{O}((x-1)^2)}{-\frac \pi2(x-1)+\mathcal{O}((x-1)^3)}\\\\ &=-\frac{2}{\pi (x-1)}+\mathcal{O}(x-1). \end{align}

The notation $$\displaystyle f(x)=\mathcal{O}((x-x_0)^n)$$ means that there exists some constant $C$ around $x_0$ such that $$\displaystyle |f(x)|\leq C|x-x_0|^n$$ for all $x$ in a neighbourhood of $x_0$.

• Thank you. My main question is about the limit of $\tan \left(x \pi/2 \right)$ (the rest in easy). I can't use the first equation (otherwise I must prove it first). Commented May 9, 2015 at 17:02
• @user239211 Ok, let me add some details. Thanks. Commented May 9, 2015 at 17:03
• @OlivierOloa I am just curious, what does the O(x-1) term represent? Is it the rest of the taylor expansion of tan(pi/2 x)? Thanks. Commented May 9, 2015 at 17:07
• Thank you @OlivierOloa . Great proof. Commented May 9, 2015 at 17:26
• @user239211 You are welcome! Commented May 9, 2015 at 17:30

For $\;x\;$ pretty close to $\;1\;$ from the right,

$$\;\lfloor x\rfloor=1\implies \tan\dfrac{\pi x}2\lfloor x\rfloor=\tan\dfrac{\pi x}2\to-\infty\;$$

and from $\;x\;$ very close to $\;1\;$ from the left you get

$$\lfloor x\rfloor=0\;\implies \tan\dfrac{\pi x}2\lfloor x\rfloor=0\to 0$$

• Please read my question again. The whole point is proving that $\tan \frac{\pi x}{2}$ goes to $\pm \infty$ (depends on the side) using the most basic tools. Commented May 9, 2015 at 17:05
• No. I said that I can use $\lim_{x \to \frac{\pi}{2}^+}\tan x=-\infty$. I know that this is the same as $\lim_{x \to 1+}\tan \left( \frac{\pi x}{2} \right)$ but I must prove it first. Until then these are, formally, different limits. Commented May 9, 2015 at 17:09
• don't you see that $\lim_{x \to \frac{\pi}{2}^+}\tan x$ and $\lim_{x \to 1+}\tan \left( \frac{\pi x}{2} \right)$ are formally different limits? Can't you understand that you must prove that they are the same? That's how math works. You must prove things. You didn't prove anywhere that $\tan\dfrac{\pi x}2\to-\infty\;$. You just said that it is a "rather well known trigonometric function" and completely ignored my question. Yes, I know that intuitively these are the same limits but intuition is not a proof. I'm sorry for you that you don't understand that. Commented May 9, 2015 at 17:25
• @user239211 Whatever: that's not the way I studied it nor the way I'd teach it. For me, one of the first things to do with limits is to show that if the limit exists then it exists no matter how the variable approaches the desired point. That is exactly what you have here, when instead of having $\;\lim_{x\to\pi/2}\tan x\;$ you have $\;\lim_{x\to 1}\tan\frac{\pi x}2\;$ . No intuition! in particular that other answer uses Taylor expansions' approximations, which are waay after limits and derivatives are studied. It doesn't matter, though: you've been answered to your satisfaction. Commented May 9, 2015 at 17:55
• @Lucas - I was very nice to him even though he didn't answer my question. But then he decided to argue with me about the limit being "easy", "and rather well known". He didn' even consider that maybe my calculus course is a bit different (in order of theorems). He didn't even read my question. I was asking how to prove $\lim_{x \to 1^+} \tan \left(\frac{\pi x}{2} \right ) = -\infty$. He didn't prove it. Instead, he showed me the "answer" which I actually already knew (he assumed that $\lim_{x \to 1^+} \tan \left(\frac{\pi x}{2} \right ) = -\infty$). Commented May 10, 2015 at 9:02