1

$$\lim_{\large x \to 1^+} \frac{2x+3}{x-1}=∞$$

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.

Absurd
  • 369
  • 1
  • 14
  • For $x>1$ You have $\lfloor x\rfloor =1$ and for $0<x<1$ You have $\lfloor x\rfloor =0$. In both cases You arrive at the limit $-\infty$. The first part contains some inaccuracies. – Peter Melech Oct 28 '19 at 14:14
  • Of course I meant: For $1<x<2$ it is $\lfloor x\rfloor =1$, which is all You need since You are interested in neighborhoods of $1$ only – Peter Melech Oct 28 '19 at 14:24
  • Beware in the first part, $\delta < \frac{7}{M}$ is necessary but not sufficient! – Olivier Roche Oct 28 '19 at 14:53
  • yes that's right, actually I said $\delta$ is less than or equals to min$(1,\frac{7}{M})$ – Absurd Oct 28 '19 at 15:11

2 Answers2

1

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 $$\lim_{x\to 1}\frac{(-1)^{\lfloor x\rfloor}}{x-1}=-\infty.$$

Sumanta
  • 9,534
  • 1
    that's right,can you please also tell me that my answer is right or not? – Absurd Oct 28 '19 at 14:38
  • Yes, your answer is right. – Sumanta Oct 28 '19 at 14:39
  • there is something about $0<x<1$,consider if $\large \frac{1}{1-x}<-M$ then $1-x>\frac{1}{-M}$, but about $1\le x<2$ we have $\frac{-1}{1-x}<-M$ so $1-x<\frac{1}{M}$,finally we can take $\delta\le\frac{1}{M}$ which is true, can you tell me why for the first case we don't have such thing? – Absurd Oct 28 '19 at 19:14
0

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. $1 < x <2$, then $(-1)^{\lfloor x \rfloor} = -1$ and $\frac{1}{M}>x-1>0$ hence $\frac{(-1)^{\lfloor x \rfloor}}{x-1} < -M$
  2. $0 < x <1$, then $(-1)^{\lfloor x \rfloor} = 1$ and $-\frac{1}{M}<x-1<0$ hence $\frac{(-1)^{\lfloor x \rfloor}}{x-1} < -M$
Olivier Roche
  • 5,319
  • 9
  • 16