Page 13 - Calc for the Clueless

Given ε > 0, if we take any x value such that 0 < |x - a| < δ, the interval on the x axis, and find the
corresponding y = f(x) value, this y value must be within a of L, that is, |f(x) - L| < ε.

Note
The point (a, L) way on way not be actually there, depending on the function. But remember, we
are only interested in values of x very, very close to a but not exactly at a.

Take a smaller ε 2; we can find δ 2 such that 0 < |x - a| < δ 2, If(x) - L| < ε 2. The smaller the ε, the smaller the δ.
f(x) goes to L as x goes to a.

Although this definition is extremely difficult, its application is pretty easy. We need to review six facts, four
about absolute value and two about fractions.
1. |ab| = |a| |b|

2. .

3 |a - b| = |b - a|.

4 |a + b| < |a| + |b|.

5. In comparing two positive fractions, if the bottoms are the same and both numerators and denominators are
positive, the larger the top, the larger the fraction. 2/7 < 3/7.

6. If the tops are the same, the larger the bottom, the smaller the fraction. 3/10 > 3/11.

Now let's do some problems.

Example 18—

Using ε, δ, prove

In the definition

f(x) = 4x - 3, a = 2, L = 5. Given ε > 0, we must find δ > 0, such that if 0 < |x-2| < δ, |(4x-3)-5| < ε.

Example 19—

Prove