How to prove $\lim_{x\to 1^+} \frac{x}{x-1} = +\infty$ with the limit definition? 
By using the definition of limit ONLY, prove that
  $$
\lim_{x\to 1^+} \dfrac{x}{x-1} = +\infty
$$
  (Note that you need to use the definition of one-sided limit from the right and you are NOT allowed to use any algebra or any theorem for limit.)

I keep getting mixed up with the definitions when trying to answer this, when attempting the question i've started with 
For all $M\in\mathbb{R}$ we need to find a $\delta>0$ such that $\dfrac{1}{1-x}>M$ for all $x\in\mathbb{R}$ and $0<x<\delta$
Not too sure what to do from here or if this is even correct.
 A: hint
$$\frac {x}{x-1}=1+\frac {1}{x-1} $$
We need prove that
$$(\forall A>2 )\;\;\;(\exists\eta>0)\;\;(\forall x>1)$$
$$0 <x-1 <\eta \implies 1+\frac {1}{x-1}>A $$
this last condition is equivalent to
$$ x-1 <\frac {1}{A-1}.$$
So we can take  $\eta=\frac {1}{A-1} $.
A: Set $y = x-1$,  then:
$\lim_{y \rightarrow 0^+} \frac{y+1}{y} = + \infty$, or
$\lim_{y \rightarrow 0^+} ( 1 + \frac{1}{y}) = + \infty$.
Let $M \in \mathbb{R^+}$ be given. 
Choose  $0 \lt \delta \lt 1/M$.
For $0 \lt y \lt \delta$: 
$ M \lt  1 + \frac{1}{y}$.
A: The actual definition of $\lim_{x \rightarrow 1^+}\frac x{x-1} = +\infty$ is actually:
For any $M \in \mathbb R$ one can find a $\delta$ (probably dependent upon the value of $M$) so that whenever $0 < |x- 1| < \delta$ and $x > 1$ then $\frac x{x-1} > M$.
This is different than what you wrote. 
Use algebra to solve if $x-1 < ????what???$ and $x > 1$ then $\frac x{x-1} > M$.
Hint:  If $x-1< \delta$ then $\frac x{x-1} > \frac x {\delta} > \frac 1{\delta}$.
So when is $\frac 1{\delta} > M$?
A: Note that 
$$
\frac{x}{x-1} > \frac{1}{x-1}
$$
if $0 < x-1 < 1$. Given any $M > 0$, we have 
$$
\frac{1}{x-1} > M
$$
if in addition $0 < x-1 < 1/M$.
So $0 < x-1 < \min \{ 1, 1/M \}$ implies $\frac{x}{x-1} > M$. Taking $\delta := \min \{ 1, 1/M \}$ suffices.
