How do you prove that the limit of #5x^2 =5# as x approaches 1 using the epsilon delta proof?

1 Answer
Jul 27, 2018

Evaluate the difference:

#abs(5x^2-5) = 5abs(x^2-1)#

#abs(5x^2-5) = 5abs(x-1)abs(x+1)#

Now for any #epsilon > 0# choose #delta_epsilon < min(1,epsilon/15)#.

As #delta_epsilon < 1# then for #x in (1-delta_epsilon, 1+delta_epsilon)# we have that:

#0 < x < 2#

and:

#1 < 1+x < 3#

Then #abs(1+x) = (1+x) < 3# and so:

#abs(5x^2-5) = 5abs(x-1)abs(x+1) < 15abs(x-1)#

On the other hand, as #delta_epsilon < epsilon/15# then for #x in (1-delta_epsilon, 1+delta_epsilon)# we have that:

#abs(x-1) < delta_epsilon < epsilon/15#

So:

#abs(5x^2-5) < 15abs(x-1)#

#abs(5x^2-5) < 15epsilon/15#

#abs(5x^2-5) < epsilon#

which proves the limit.