Page 14 - Calc for the Clueless

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Given ε > 0, we must find δ > 0, such that if 0 < |x-4| < δ, |x +2x-24| < ε.

We must make sure that |x + 6| does not get too big. We must always find δ, no matter how small. We must take
a preliminary δ = 1. |x - 4| < 1, which means-l < x - 4 < 1 or 3 < x < 5. In any case x < 5. Sooooo...

Finishing our problem, |x + 6| |x- 4| < 11 • δ = ε δ = minimum (1, ε/11).

Example 20—

Prove

Again take a preliminary δ = 1. |x - 5| < 1. So 4 < x < 6. To make a fraction larger, make the top larger and the
bottom smaller. 0 < |x - 5| < δ. We substitute δ on the top. Since x > 4, we substitute 4 on the bottom.

If δ = minimum (1, 10ε), |2/x - 2/5| < ε.

Continuity

We finish with a brief discussion of continuity of a function at a point. Intuitively, continuity at a point means
there is no break in the graph of the function at the point. Let us define continuity more formally:

f(x) is continuous at point a if

1.

2.

We will do a longish example to illustrate the definition fully.

Example 21—

Let

We wish to examine the continuity at x = 1, 3, and 6.