Question

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) (-1/(x-2)^2) =-\infty`

I wrote,

The limit exists `lim_(x to 2) (-1/(x-2)^2) =-\infty` if for all B < 0, exists a `\delta`, such that `-1/(x-2)^2` < B, always that 0 < |x-2| < `\delta`.
Looking for inequality we can choose the `\delta` more appropriate.

`-1/(x-2)^2 < B`
`-(x-2)^2 > 1/B`

I'm stuck here because I need the `\delta` positive. I don't know, how I complete this proof.

Answer 1

See below. You can always choose for instance `delta=1/2sqrt(-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`, `(-1/B)>0`
So if `-δ^2>1/B` => `delta^2<-1/B`
or `delta < sqrt(-1/B)` (remember that both `-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.