How do you prove the statement lim as x approaches 0 for #x^2 = 0# using the epsilon and delta definition?

1 Answer
Jim H
Oct 9, 2015

See the explanation, below.

Explanation:

You need to show that if we are given a positive number that we'll call #epsilon#, then there is a number #delta# (also positive) that makes the following true:

if #x# is chosen so the #0 < abs(x-0) < delta#, then #abs(x^2-0) < epsilon#.

One way to do this is to notice that if #absx < sqrt(epsilon)#, the #absx < epsilon#

Choose #delta = sqrt epsilon#

Another way is to observe that for #absx < 1#, we have #abs(x^2) < absx#. Choose #delta = min {1,epsilon}#

Whichever way is chosen, we then write up the proof:

Proof

Given #epsilon > 0#, choose #delta = "whatever"#

Now if #0 < abs(x-0) < delta#, then

[insert a proof that #absx < 0# implies #abs(x^2) < epsilon#]

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