How do you prove the statement lim as x approaches 4 for $(x^2) = 4$ using the epsilon and delta definition?

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Answer 1 · 2015-10-15T11:17:32

Note that the limit as x approaches 4 is 16, and the limit as x approaches 2 is 4.
I will prove the latter case for you.

By definition, $lim_(x->x_0)f(x)=Liff AA epsilon>0, EE delta >0$ such that $|x-x_0| < delta =>|f(x)-L| < epsilon$

Let $epsilon > 0$ and select $k in RR^+$ such that $|x-2|< k$

$therefore -k < x-2 < k$

$therefore 4-k < x+2 < k+4$

Hence $|x+2| < k+4$

Now select $delta=min{epsilon/(k+4); k}$

Clearly $delta>0$

Let $|x-2| < delta$

Now : $|f(x)-4|=|x^2-4|$
$=|x-2|*|x+2|$

$< delta*epsilon/delta$

$=epsilon$

This then proves that $lim_(x->2)x^2=4$

How do you prove the statement lim as x approaches 4 for (x^2) = 4 (x^2) = 4 using the epsilon and delta definition?