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Use the Epsilon Delta Method to Prove That Any Linear Function is Continuous

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Using the Epsilon Delta Definition of a Limit

We can use the epsilon-delta definition of a limit to confirm some "expectation" we might have for the value some expression "should have had" when one of its variables takes on some value -- were it not for some pesky numerator and denominator becoming zero, or some similar problem happening at just the wrong moment.

However, before dealing with more problematic expressions -- let us see how this definition plays out in a tamer situation.

Indeed, let us focus just on a piece of this definition first. The epsilon-delta definition works a bit like a challenge/response. For a given $\epsilon \gt 0$, we are challenged to find a $\delta \gt 0$ that satisfies the given criteria.

Consider the function $f(x) = 4x+1$. Clearly, at $x=3$ this function takes on the value of $4(3)-1 = 11$. There is nothing "blowing up" things into indeterminant form here. Reality and our expectations are in complete agreement.

As such, we rightly expect

$$\lim_{x \rightarrow 3} (4x - 1) = 11$$

If this is true, then we should be able to pick any $\epsilon \gt 0$, say $\epsilon = 0.01$, and find some corresponsding $\delta \gt 0$ whereby whenever $0 \lt |x - 3| \lt \delta$, we can be assured that $|f(x) - 11| \lt 0.01$.

Let's put that to the test.

Starting with the last expression, let us replace $f(x)$ with $(4x-1)$:

$$\begin{array}{rcl} |f(x) - 11| &=& |(4x-1) - 11|\\ &=& |4x-12|\\ &=& 4|x-3| \end{array}$$

The above makes it clear that $|f(x)-11|$ stays four times the size of $|x-3|$. (This is of course not surprising, as $f(x)$ was a linear function with slope $4$.)

However, what this means is that if we want to keep $|f(x)-11| \lt 0.01$, we'll need to keep $|x-3|$ four times smaller.

Thus, if we take $\delta$ to be any value less than $0.0025$, then $$|f(x) - 11| \lt 0.01 \quad \textrm{ whenever } 0 \lt |x - 3| \lt \delta$$

The above calculation suggests what we might do for any other value of $\epsilon$ -- we simply use a $\delta$ value one-fourth the size of the $\epsilon$ in question to ensure the needed inequality to establish the limit.

Here again, what we require in order to say $\lim_{x \rightarrow 3} (4x-1) = 11$ is that for any $\epsilon \gt 0$, we can find a $\delta \gt 0$ such that if $0 \lt |x - 3| \lt \delta$, then $|f(x) - 11| \lt \epsilon$.

So, letting $\delta = \epsilon / 4$, whenever $0 \lt |x - 3| \lt \delta$ we have

$$\begin{array}{rcl} |x-3| &\lt& \epsilon/4\\ 4|x-3| &\lt& \epsilon\\ |4x-12| &\lt& \epsilon\\ |(4x-1) - 11| &\lt& \epsilon\\ |f(x) - 11| &\lt& \epsilon \end{array}$$

which was the desired result.

As such, we can definitively say as a consequence of the epsilon-delta definition of a limit that

$$\lim_{x \rightarrow 3} (4x-1) = 11$$

The expression $4x-1$ in the last example was a linear one, and led to a $\delta$ that could be used in the definition which was really a very simple function of $\epsilon$. This is more the exception than the rule. As such, let us look at a slightly more complicated example...

Suppose we wish to show that $$\lim_{x \rightarrow 2} x^2 = 4$$

According to the definition, we will thus need to show that for any $\epsilon \gt 0$, we can find a $\delta \gt 0$ such that $$|x^2 - 4| \lt \epsilon \quad \textrm{ whenever } \quad 0 \lt |x-2| \lt \delta$$

Equivalently, whenever $0 \lt |x-2| \lt \delta$, we want

$$\begin{array}{rcl} |(x-2)(x+2)| &\lt& \epsilon\\ |x-2||x+2| &\lt& \epsilon\\ |x-2| &\lt& \displaystyle{\frac{\epsilon}{|x+2|}} \end{array}$$

At this point, one may wish to try to pick $\delta = \epsilon / |x+2|$, but we must be careful. This expression involves $x$, and $\delta$ should only depend on the value of $\epsilon$, according to the definition.

Recall that $\delta$ creates a band around $c=2$, so that when $x$ is chosen from this band (with the possible exception of $x=c$), we hope to keep $f(x)$ within $\epsilon$ of the limiting value $4$.

Let's consider a fixed $\delta$-band about $c=2$, say where $\delta = 1$. Within this band, we only consider $x$ values in the interval $(1,3)$.

Note in particular what happens to the expression $\displaystyle{\frac{\epsilon}{|x+2|}}$ at the ends of this interval: $x = 1$ and $x = 3$.

Pushing this idea a little further, we can say the following for any $x$ in this band where $|x-2| \lt 1$, $$\frac{\epsilon}{5} \lt \frac{\epsilon}{|x+2|}$$

So here's the trick -- let us pick $\delta$ to be the smaller of $1$ and $\epsilon / 5$. In this way, $\delta$ does not depend on any specific $x$ -- and only on $\epsilon$.

Further, consider the two cases for $\epsilon$ that can now result:

  1. If $\epsilon / 5 \gt 1$, then we take $\delta = 1$ and observe that whenever $0 \lt |x-2| \lt \delta$, $$|x-2| \lt 1 \lt \epsilon / 5 \lt \frac{\epsilon}{|x+2|}$$

  2. If $\epsilon / 5 \le 1$, then we take $\delta = \epsilon / 5$ and observe that whenever $0 \lt |x-2| \lt \delta$, $$|x-2| \lt \epsilon / 5 \le 1 \quad \textrm{ and } \quad |x-2| \lt \epsilon / 5 \lt \frac{\epsilon}{|x+2|}$$

Recall that $\displaystyle{|x-2| \lt \frac{\epsilon}{|x+2|}}$ was equivalent to $|x^2 - 4| \lt \epsilon$ by our earlier argument.

Thus, we have shown that for any $\epsilon \gt 0$, we can find a $\delta \gt 0$ (namely, the minimum value in $\{1,\epsilon/5\}$) so that $$|x^2-4| \lt \epsilon \quad \textrm{ whenever } \quad 0 \lt |x-2| \lt \delta$$

Having satisfied the epsilon-delta definition of a limit, we can then say with confidence $$\lim_{x \rightarrow 2} x^2 = 4$$

Whew! That was a lot of work to establish our expectation for the value of $x^2$ when $x=2$. Just like the linear function we considered earlier, there was nothing weird that was going to happen to $x^2$ upon attempting to evaluate this expression at $x=2$. No indeterminant form or other problem results.

So why do all this work? And Lord help us if something weird did happen!

Using the epsilon-delta definition to establish limits can clearly be tricky and complicated. However, it is the bedrock upon which we build the notion of limits and can consequently not be completely ignored.

However, it can be somewhat kept hidden from view.

One might be worried about what happens when looking at limiting values of more complicated expressions -- expressions that are combinations (e.g., sums, differences, products, quotients, etc) of simpler expressions, like the linear expression we examined, and other primitive expressions (e.g., $n^{th}$ powers, sines, logs, exponentials, etc).

For example, if finding a limiting value of $x^2$ gave us grief -- imagine trying to find a limiting value for the expression $x^2 - 5x^3 +\sin \pi x$!

Thankfully, we can prove -- using the epsilon-delta definition -- both of the following:

  1. straight-forward results about how to find limits of simple functions (like the linear $4x-1$ with which we started, or $x^2$, or $\sin x$, etc), and

  2. how to find limits of combinations of expressions (e.g., sums, differences, products, etc) from the limiting values of their individual parts.

Once these results have been proven, we can turn our attention to their application to find limiting values rather than always going "back to the well" of the epsilon-delta definition to prove things "from scratch".

As you will soon see, applying these results to find limiting values will be substantially easier than using the epsilon-delta definition -- but don't lose sight of the fact that these results employed only work because of the epsilon-delta definition. We can't simply say, "Oh we've found an easier way -- that epsilon-delta technique was such a waste!" It doesn't work like that. Both ideas are crucial in their own way.

Interestingly, this technique of accomplishing some task in a very complicated way initially, only to then prove useful properties about simple functions and combinations of functions that will let us largely bypass the more complicated method will be a recurring theme throughout our study of calculus -- but more on that later...

grey-smithcronts.blogspot.com

Source: http://math.oxford.emory.edu/site/math111/epsilonDeltaExamples/