Example 2

In the limit

lim_x→2 x² = 4,

find a δ > 0 such that whenever 0 < |x - 2| < δ, |x² - 4| < 1/10. By the Lemma, we must find δ > 0 such that whenever

2 - δ < x < 2 + δ and x ≠ 2,

4 - 1/10 < x² < 4 + 1/10

Assume that

2 - δ < x and x < 2 + δ.

As long as 2 - δ and x are positive we may square both sides,

4 - 4δ + δ² < x² and x² < 4 + 4δ + δ²

4 + (-4δ + δ²) < x² and x² < 4 + 4(δ + δ²).

Now take δ small enough so that

-1/10 ≤ -4δ + δ² and 4δ + δ² ≤ 1/10 .

For example, δ = 1/50 will do. Then

4 - 1/10 < x² < 4 + 1/10.

Thus whenever

0 < |x - 2| < 1/50, |x² - 4| < 1/10.

Notice that any smaller value of δ, such as δ = 1/100, will also work.