# My proof for this limit using the definition is correct? lim to 2^+ (1/(x-2)) = +\infty

## My answer: For all A > 0, exists $\setminus \delta$ > 0 such that: $\left(\frac{1}{x - 2}\right) > A$ so that 0 < x+2 < $\setminus \delta$. Looking on inequality bellow between B, we have the key choose for $\setminus \delta$ : $\left(\frac{1}{x - 2}\right) > A$ $\left(x - 2\right) < \frac{1}{A}$ $x < \frac{1}{A} + 2$ Like this, for $\setminus \delta$ = $\frac{1}{A} + 2$, we have $\frac{1}{x - 2} > A$ always that 0 < x-2 < $\delta$.

May 31, 2018

See explanation

#### Explanation:

There is one mistake: $0 < x + 2 < \delta$. After all, it is $x - 2$ that you want to go towards 0.

I might also want to refine the wording a little, for instance:
"For all A > 0, there exists a $\delta > 0$ such that:" to make the proof clearer.

Also, as the proof presupposes that $x > 2$, I might write:
$2 < x < 2 + \frac{1}{A}$ to make it clear that x lies between 2 and $2 + \frac{1}{A}$.

One other detail: You introduce B, but it's not clear where B belongs or what it refers to.