Limit by epsilon-delta proof: Example 2

November 1, 2011 GB Calculus and Analysis, Problem Solving and Proofs

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 $|x + 2| < k$ .

If $x$ is confined to some interval centered at $2$ , then we can find $k$ . For instance, suppose $|x-2| < 1$ , which is the same as $1 < x < 3$ , then $3 < x+2 < 5$ . In particular $|x + 2| = x + 2 < 5$ .

Therefore,

$|x+2||x-2| < 5|x-2| <\epsilon$ . Hence, we have $|x - 2| < \frac{\epsilon}{5}$ . But we have two restrictions:

$|x - 2 | < 1$ and $|x-2| < \frac{\epsilon}{5}$

so, to be sure that both inequalities are obeyed, we choose $\delta$ to be smaller than $1$ and $\frac{\epsilon}{5}$ abbreviated as $\delta \leq \min (1, \frac{\epsilon}{5})$ whenever $|x^2 - 4| < \epsilon$ .

Limit by epsilon-delta proof: Example 2

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