How do you use the epsilon-delta definition of continuity to prove #f(x) = x^2# is continuous?

1 Answer
Dec 15, 2016

Answer:

For any #epsilon >0# we chose #delta# such that:

#delta (2|barx|+delta) < epsilon#

Explanation:

By definition, #f(x)# is continuous for #x=barx#, if for any real number #epsilon>0# we can find a real number #delta >0# such that:

#x in (barx-delta, barx+delta) => |f(x)-f(barx)| < epsilon#

Now, for #x in (barx-delta, barx+delta)#, we have that:

#|x-barx|< delta#,
#|x|<|barx|+delta#

and then:

#|f(x)-f(barx)| = |x^2-barx^2| = |(x-barx)(x+barx)|= |x-barx||x+barx| <=delta (|x|+|barx|) <= delta (2|barx|+delta) #

So, for any #epsilon >0# if we chose #delta# such that:

#delta (2|barx|+delta) < epsilon#

The condition of continuity is satisfied.