# How I finish this proof using the definition of limit for this lim_(x to 2) (-1/(x-2)^2) =-\infty ?

## ${\lim}_{x \to 2} \left(- \frac{1}{x - 2} ^ 2\right) = - \setminus \infty$ I wrote, The limit exists ${\lim}_{x \to 2} \left(- \frac{1}{x - 2} ^ 2\right) = - \setminus \infty$ if for all B < 0, exists a $\setminus \delta$, such that $- \frac{1}{x - 2} ^ 2$ < B, always that 0 < |x-2| < $\setminus \delta$. Looking for inequality we can choose the $\setminus \delta$ more appropriate. $- \frac{1}{x - 2} ^ 2 < B$ $- {\left(x - 2\right)}^{2} > \frac{1}{B}$ I'm stuck here because I need the $\setminus \delta$ positive. I don't know, how I complete this proof.

May 10, 2018

See below. You can always choose for instance $\delta = \frac{1}{2} \sqrt{- \frac{1}{B}}$ regardless of B.

#### Explanation:

0 < |x-2| < δ => 0 < (x-2)^2 < δ^2

So if -δ^2>1/B it follows that
-(x-2)^2> -δ^2>1/B

As $B < 0$, $\left(- \frac{1}{B}\right) > 0$
So if -δ^2>1/B => ${\delta}^{2} < - \frac{1}{B}$
or $\delta < \sqrt{- \frac{1}{B}}$ (remember that both $- \frac{1}{B}$ and $\delta$ are positive.)

This can always be fulfilled, since you for any B can choose for instance
delta=1/2sqrt(-1/B) < sqrt(-1/B))

I hope this helps you on your way to solve your proof.