How do you use the epsilon delta definition to prove that the limit of #sqrtabsx=0# as #x->0#?

Question

1300 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-09T20:25:58

See explanation below.

For any #epsilon > 0# choose #delta_epsilon < epsilon^2#.

Now, for #x in (-delta_epsilon , delta_epsilon )# we have:

#abs (x) < delta_epsilon < epsilon^2#

and as # sqrt(x)# is a strictly increasing function:

#sqrt(abs(x)) < sqrt(epsilon^2)#

or:

#sqrt(abs(x)) < epsilon#

as #sqrt(x)# is positive, we also have:

#abs(sqrt(x)-0) < epsilon#

which proves that:

#lim_(x->0) sqrt(abs(x)) = 0#