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.

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.


|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.

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2 thoughts on “Limit by epsilon-delta proof: Example 2

  1. Hi Guillermo Bautista.I’m a 10th grader,and actually practicing Calculus,thus I’m working on Integration by Partial Fractions,doing great.But I felt like I needed to go back a little on Limits and Continuity(though I’m good at it) for the intriguing epsilon-delta definition.I was just wondering if you had some good examples,explanations and proofs about how to deal with non-linear functions(let it be polynomial,logarithmic,exponential,rational and trigonometric functions).Here is my appreciate your help.

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