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?
 A: 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 $x_0$.

A: You already have it...almost ! 
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 
$$\lfloor x\rfloor=0\;\implies \tan\dfrac{\pi x}2\lfloor x\rfloor=0\to 0$$
