Category Archives: Calculus and Analysis

Limit by epsilon-delta proof: Example 2

Posted on November 1, 2011 by Guillermo Bautista | Leave a comment

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 $|x + 2| < k$ . Continue reading →

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Posted in Calculus and Analysis, Problem Solving and Proofs

Tagged calculus limits, epsilon-delta definition, limits by definition, proof of limits by definition

Limit by epsilon-delta proof: Example 1

Posted on February 27, 2011 by Guillermo Bautista | 14 Comments

We have discussed extensively the meaning of the $\epsilon-\delta$ definition. In this post, we are going to learn some strategies to prove limits of functions by definition. The meat of the proof is finding a suitable $\delta$ for all possible $\epsilon$ values.

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 $|f(x) - L| < \epsilon$ .

Example 1: Let $f(x) = 3x + 5$ . Prove that $\lim_{x \to 2} f(x) = 11$

If we are going to study definition limit above, and apply it to the given function, we have $\lim_{x \to 2} 3x + 5 = 11$ , if for all $\epsilon > 0$ , however small, there exists a $\delta > 0$ such that if $0 < |x - 2| < \delta$ , then $|3x + 5 - 11| < \epsilon$ . We want to find the value of $\delta$ , in terms of $\epsilon$ ; therefore, we can manipulate one of the inequalities to the other’s form. In particular, we will manipulate $|3x + 5 - 11| < \epsilon$ to an expression such that the expression inside the absolute value sign will become $x - 2$ .

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Posted in Calculus and Analysis, College Mathematics

Tagged epsilon-delta definition, epsilon-delta explanation, epsilon-delta problems, epsilon-delta proof, epsilon-delta proof example

Constructing segments with irrational lengths

Posted on October 25, 2010 by Guillermo Bautista | Leave a comment

Since rational numbers can be expressed as fractions (or ratio of two integers), rational numbers can be easily located on the number line using compass and straightedge construction. In principle, a segment of any length can be divided into any number of parts; hence, it is possible to locate any rational number on the number line. Irrational numbers, on the other hand, cannot be expressed as ratio of two integers, so the big question is:

How do we locate irrational numbers on the number line?

The question above is equivalent to: “How do we construct a segment with irrational lengths?”

It is easy to locate some irrational numbers on the number line even with compass and straightedge construction. The irrational $\sqrt{2}$ can be located by constructing square ABCD (Figure 1), getting the diagonal AC, and constructing a circle with radius AC. It follows that the length of AE is $\sqrt{2}$ , and the coordinates point E are $(\sqrt{2},0)$ .

Figure 1

It is also apparent that the easiest way to construct segments with irrational lengths is by constructing diagonals of rectangles. In Figure 2, OB, OC and OD have lengths $\sqrt{2}$ , $\sqrt{5}$ , and $\sqrt{10}$ respectively. More complicated construction is required to construct other irrational lengths, $\sqrt{3}$ for instance, which is the length of EH. Continue reading →