Limit by epsilon-delta proof: Example 2

This is the overdelayed continuation of the discussion on the \epsilon-\delta definition of limits. In this post, we discuss another example.

Prove that the \lim_{x \to 2} x^2 = 4.

Recall that the definition states that the limit of f(x) = L as x approaches a if for all \epsilon > 0, however small, there exists a \delta > 0 such that if 0 < |x - a| < \delta, then |f(x) - L| < \epsilon.

From the example 1, we have learned that we should manipulate |f(x)-L=|x^2 - 4|, to make one of the expressions look like |x-a|=|x-2|. Solving,  we have

|f(x) - L| = |x^2 - 4| = |(x+2)(x-2)| = |x+2||x-2|.

Note that we have accomplished our goal, going back to the definition, this means that if 0 < x - 2 < \delta, then |x+2||x-2| < \epsilon.

Now, it is not possible to divide both sides by x + 2 (making it |x-2| < \frac{\epsilon}{|x+2|}) because x varies. This means that we have to find a constant k such that |x + 2| < k. » Read more