calculus - 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?

Answer

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 $x \to 1$.

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

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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 $x_0$.