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.

1 Answer
May 10, 2018

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.