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.