Please explain step by step how you came to your conclusion. High rating will be given for best answer. Thanks so much!!
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By the ε-δ definition of a limit, lim (x-->c) f(x) = L means that for each ε > 0 there exists δ > 0 such that:
|f(x) - L| < ε whenever 0 < |x - c| < δ.
So, we wish to show that for each ε > 0 there exists δ > 0 such that:
|(x^2 - 2x + 4) - 7| < ε whenever 0 < |x - 3| < δ
==> |x^2 - 2x - 3| < ε whenever 0 < |x - 3| < δ.
Notice that we can factor x^2 - 2x - 3 as (x - 3)(x + 1), so:
|x^2 - 2x - 3| = |(x - 3)(x + 1)| = |x - 3||x + 1|.
Then:
|x - 3||x + 1| < ε ==> |x - 3| < ε/|x + 1|.
Of course, we cannot x's involved in our choice of δ. To get around this, it is sufficient to just choose a region around x = 3, like 2 < x < 4, to prove continuity. For 2 < x < 4, 3 < |x + 1| < 5, so:
|x - 3| < ε/5,
which suggests that we choose δ = min{1, ε/5}.
(Note that both |x - 3| < 1 and |x - 3| < ε/5 need to be satisfied, so we choose the smaller of the two values.)
At this point, for each ε > 0 choose δ = min{1, ε/5}. To show that this choice works, we need to check both δ = 1 and δ = ε/5. If δ = 1, then:
|x - 3| < 1 ==> 2 < x < 4 ==> 3 < |x + 1| < 5.
If we selected 1 over ε/5, then 1 < ε/5 ==> 5 < ε. Since |x - 3| < 1 and |x + 1| < 5:
|x - 3| < 1 ==> |x + 1||x - 3| < |x - 3| < 5 < e, as required.
We know that if ε/5 is selected, then ε/5 < 1 ==> ε < 5. Hence:
|x - 3| < ε/5 ==> |x + 1||x - 3| = ε|x + 1|/5 < e(5)/5 = e,
which concludes the proof.
I hope this helps!
|f(x) - L| < ε whenever 0 < |x - c| < δ.
So, we wish to show that for each ε > 0 there exists δ > 0 such that:
|(x^2 - 2x + 4) - 7| < ε whenever 0 < |x - 3| < δ
==> |x^2 - 2x - 3| < ε whenever 0 < |x - 3| < δ.
Notice that we can factor x^2 - 2x - 3 as (x - 3)(x + 1), so:
|x^2 - 2x - 3| = |(x - 3)(x + 1)| = |x - 3||x + 1|.
Then:
|x - 3||x + 1| < ε ==> |x - 3| < ε/|x + 1|.
Of course, we cannot x's involved in our choice of δ. To get around this, it is sufficient to just choose a region around x = 3, like 2 < x < 4, to prove continuity. For 2 < x < 4, 3 < |x + 1| < 5, so:
|x - 3| < ε/5,
which suggests that we choose δ = min{1, ε/5}.
(Note that both |x - 3| < 1 and |x - 3| < ε/5 need to be satisfied, so we choose the smaller of the two values.)
At this point, for each ε > 0 choose δ = min{1, ε/5}. To show that this choice works, we need to check both δ = 1 and δ = ε/5. If δ = 1, then:
|x - 3| < 1 ==> 2 < x < 4 ==> 3 < |x + 1| < 5.
If we selected 1 over ε/5, then 1 < ε/5 ==> 5 < ε. Since |x - 3| < 1 and |x + 1| < 5:
|x - 3| < 1 ==> |x + 1||x - 3| < |x - 3| < 5 < e, as required.
We know that if ε/5 is selected, then ε/5 < 1 ==> ε < 5. Hence:
|x - 3| < ε/5 ==> |x + 1||x - 3| = ε|x + 1|/5 < e(5)/5 = e,
which concludes the proof.
I hope this helps!
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I've no idea what "delta/epsilon definition" means, but the limit as x>>3 is obtained by substituting 3 for x
You thus get 3^2 - 2*3 + 4 .....9 - 6 + 4 ... 7
You thus get 3^2 - 2*3 + 4 .....9 - 6 + 4 ... 7