Prove the statement using the epsilion delta definition of limit.
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I will give a proof first, and then explain where it comes from.
Let epsilon>0. Set delta=min{1, epsilon/8}.
Suppose that 0<|x-3|
-delta
7-delta0, which implies that x+4 is positive. So |x+4|=x+4<7+delta< or =7+1=8. Thus
|x^2+x-4-8|=
=|x^2+x-12|
=|(x+4)(x-3)|
=|x+4||x-3|
<8|x-3| (since |x+4|<8)
<8*epsilon/8 (since |x-3|
=epsilon.
Explanation:
We need to have |x^2+x-4-8| small when x is close to 3.
|x^2+x-4-8|=|x^2+x-12|=|x+4||x-3|. The |x-3| is no problem, because we get to control how small this is. But figuring out how to deal with |x+4| presents a little bit of trouble. To relate x+4 to x-3, start with
-delta
and ask, "What could we add through the inequality to get x+4 in the middle?" Adding 7 works. This gives
7-delta
But we won't know about |x+4| unless we know that 7-delta is positive. Keeping delta from being bigger than 1 keeps delta positive. (There is nothing special about 1 here. Other choices work.) Now if delta is at most 1, then that means that 7+delta is at most 8. This is why I used epsilon/8.
I hope that this helps you.
Let epsilon>0. Set delta=min{1, epsilon/8}.
Suppose that 0<|x-3|
|x^2+x-4-8|=
=|x^2+x-12|
=|(x+4)(x-3)|
=|x+4||x-3|
<8|x-3| (since |x+4|<8)
<8*epsilon/8 (since |x-3|
Explanation:
We need to have |x^2+x-4-8| small when x is close to 3.
|x^2+x-4-8|=|x^2+x-12|=|x+4||x-3|. The |x-3| is no problem, because we get to control how small this is. But figuring out how to deal with |x+4| presents a little bit of trouble. To relate x+4 to x-3, start with
-delta
7-delta
I hope that this helps you.