using epsilon delta definition to proof limits 
$$\lim_{\large x \to 1^+} \frac{2x+3}{x-1}=∞$$

let $$f(x)=\frac{2x+3}{x-1}$$ then
$\forall M∈ℝ>0, \exists\delta>0,\forall x\in D_f \left(  
0  <    \left|x-1 \right|=x-1<\delta\Longrightarrow \large\left|\frac{2x+3}{x-1}\right|\right)>M$
$$M<\left|\frac{2x+3}{x-1}\right|=\frac{\left|2x+3\right|}{x-1}$$
take $\delta\le1$ implies:$$M<\frac{\left|2x+3\right|}{x-1}<\frac{7}{x-1}$$$$x-1<\frac{7}{M}$$
hence $$\delta\le\min\left\{1,\left(\frac{7}{M}\right)\right\}$$
is it true?

$$\lim_{\large x \to 1} \frac{\left(-1\right)^{\left[x\right]}}{x-1}=-∞$$

let $$g(x)=\frac{\left(-1\right)^{\left[x\right]}}{x-1}$$ ($\left[x\right]$ is floor function )
 then
$\forall M∈ℝ>0, \exists\delta>0,\forall x\in D_g \left( 
 0   <   \left|x-1 \right|<\delta\Longrightarrow \large\frac{\large \left(-1\right)^{\left[\large x\right]}}{x-1} <-M \right)$
take $\delta\le1$ implies:$0<x<2$, here is the problem, what is exactly $\left[x\right]$?, I tried another $\delta$ less than $1$, but still I have the problem.
 A: FIRST:--- Choose $M>0$. Let $\delta>0$ be such that,$$0<\delta<\frac{5}{M+2}$$ $$\implies\frac{5}{\delta}-2=\frac{5-2\delta}{\delta}>M.$$ Hence, for any $x\in \Bbb R$ with $1<x<1+\delta$ we have, $$\frac{2x+3}{x-1}=\frac{2(x-1)+5}{x-1}>\frac{5-2(x-1)}{x-1}>\frac{5-2\delta}{\delta}>M.$$ So, $$\lim_{\large x \to 1+} \frac{2x+3}{x-1}=∞.$$
SECOND:--- Choose $M>0$. Then for any $0<\delta<\frac{1}{M}$ we have, $$1-\delta<x<1\implies\frac{(-1)^{\lfloor x\rfloor}}{x-1}=\frac{(-1)^0}{x-1}=\frac{1}{x-1}<-\frac{1}{\delta}<-M$$$$\text{and}$$$$1<x<1+\delta\implies\frac{(-1)^{\lfloor x\rfloor}}{x-1}=\frac{(-1)^1}{x-1}=\frac{-1}{x-1}<-\frac{1}{\delta}<-M.$$ So $$\lim_{x\to 1}\frac{(-1)^{\lfloor x\rfloor}}{x-1}=-\infty.$$
A: The computations you show here are usefull, but they aren't what should appear in your proof, eg let's prove 
$$\lim_{\large x \to 1^+} \frac{2x+3}{x-1}=∞$$
We have to show that $\forall M>0, \ \exists \delta > 0, \ (x > 1 \textrm{ and } x-1 < \delta) \Longrightarrow \frac{2x+3}{x-1} >M$.
Take $M>0$. Your computation suggests that we can take $\delta_M := \min \{1;\frac{7}{M}\}$.
Indeed, if $1<x<1+\delta_M$, then $\frac{2x+3}{x-1} > (2x+3) \cdot \frac{M}{7}$, and knowing that $0<2x+3<7$ isn't usefull. What would be usefull here is knowing that $2x+3$ is bigger than 7, but that isn't the case. I suggest that you take $\delta_M := \min \{1;\frac{5}{M}\}$ since $2x+3 > 5$.
The floor function is defined as follow : $\lfloor x \rfloor$ is the biggest integer smaller or equal than x. In particular,
$$\lfloor x \rfloor = \left\{ \begin{matrix} 0 & \textrm{ if } 0\leqslant x <1 \\ 1 & \textrm{ if } 1\leqslant x < 2 \end{matrix} \right.$$
For the second limit, you'll hence have to distinguish the case where $0\leqslant x<1$ and the case where $1 \leqslant x < 2$. Anyway, taking $\delta_M := \frac{1}{M}$ works for any $M>1$:
Take $x$ such that $|x-1|<\frac{1}{M}$. Then $0<x<2$. Distinguish two cases :


*

*$1 < x <2$, then $(-1)^{\lfloor x \rfloor} = -1$ and $\frac{1}{M}>x-1>0$ hence $\frac{(-1)^{\lfloor x \rfloor}}{x-1} < -M$

*$0 < x <1$, then $(-1)^{\lfloor x \rfloor} = 1$ and $-\frac{1}{M}<x-1<0$ hence $\frac{(-1)^{\lfloor x \rfloor}}{x-1} < -M$
