## Real Analysis – Limits and Continuity V

The ${\epsilon}$ ${\delta}$ condition is somewhat hard to get into our heads as neophytes. On top of that the similarity of the ${\epsilon}$ ${\delta}$ definition for limit and continuity can increase the confusion and to try to counter those frequent turn of events the first part of this post will try to clarify the ${\epsilon}$ ${\delta}$ condition by means of examples.

${\epsilon}$ ${\delta}$ for Continuity —

First we’ll start things off with something really simple.

Let ${f(x)=\alpha}$ which is obviously continuous.

The gist of the the ${\epsilon}$ ${\delta}$ reasoning is that we want to show that no matter the ${\delta}$ that is chosen at first it is always possible to find an ${\epsilon}$ that satisfies Heine’s criterion for continuity.

Getting back to our function ${f(x)=\alpha}$ it is ${|f(x)-f(c)| < \delta}$. Here ${f(x)=f(c)=\alpha}$ so

{\begin{aligned} |f(x)-f(c)| &< \delta \\ |\alpha-\alpha| &< \delta \\ |0| &< \delta \\ 0 &< \delta \end{aligned}}

Which is trivially true since ${\delta > 0}$ by assumption. Hence any value of ${\epsilon}$ will satisfy Heine’s criterion for continuity and ${f(x)=\alpha}$ is continuous at ${c}$.

Since we never made any assumption about ${c}$ other than ${c \in {\mathbb R}}$ we conclude that ${f(x)=\alpha}$ is continuous in all points of its domain.

Let us now look at ${f(x)=x}$. Again we’ll look at continuity for point ${c}$ (${f(c)=c}$):

{\begin{aligned} |f(x)-f(c)| &< \delta \\ |x-c| &< \delta \end{aligned}}

The last expression is just we want at this stage since want to have something of the form ${x-c}$ (the first part of the ${\epsilon}$ ${\delta}$ criterion).

If we let ${\epsilon=\delta}$ it is ${|x-c| < \epsilon}$ and this completes our proof that ${f(x)=x}$ is continuous at point ${c}$.

And again since we never made any assumption about ${c}$ other than ${c \in {\mathbb R}}$ we conclude that ${f(x)=\alpha}$ is continuous in all points of its domain.

Now we let ${f(x)=\alpha x + \beta}$ and will see if ${f(x)}$ is continuous at ${c}$.

{\begin{aligned} |f(x)-f(c)| &< \delta \\ |\alpha x + \beta-(\alpha c + \beta)| &< \delta \\ |\alpha x -\alpha c| &< \delta \\ |\alpha||x-c| &< \delta \\ |x-c| &< \dfrac{\delta}{|\alpha|} \end{aligned}}

Hence if we let ${\epsilon=|\delta|/ |\alpha|}$ it is ${|x-c|< \epsilon}$ and ${f(x)=\alpha x + \beta}$ is continuous at ${c}$.

As a final example of Heine’s criterion of continuity we’ll look into ${f(x)=\sin x}$.

{\begin{aligned} |f(x)-f(c)| &< \delta \\ |\sin x-\sin c| &< \delta \end{aligned}}

Since we want something like ${|x-c| < g(\delta)}$ the last expression isn’t very useful to us.

In this case we’ll take an alternative approach which nevertheless works and has exactly the same spirit of what we’ve using so far.

Please look at every step I make with a critical eye and see if you can really understand what’s going on with this deduction.

{\begin{aligned} |\sin x-\sin c| &= 2\left| \cos\left( \dfrac{x+c}{2}\right)\right| \left| \sin\left( \dfrac{x-c}{2}\right)\right|\\ &< 2\left| \sin\left( \dfrac{x-c}{2}\right)\right| \end{aligned}}

Since ${x \rightarrow c}$ we know that at some point ${\dfrac{x-c}{2}}$ will be in the first quadrant. Thus

{\begin{aligned} 2\left| \sin\left( \dfrac{x-c}{2}\right)\right| &< 2\left|\dfrac{x-c}{2}\right| \\ &= |x-c|\\ &< \epsilon \end{aligned}}

Where the last inequality follows by hypothesis.

That is to say that if we let ${\epsilon=\delta}$ it is ${|x-c|<\epsilon \Rightarrow | \sin x - \sin x | < \delta}$ which is the epsilon delta definition of continuity.

${\epsilon}$ ${\delta}$ for Limits —

After looking into some simple ${\epsilon}$ ${\delta}$ proofs for continuity we’ll take a look at ${\epsilon}$ ${\delta}$ for limits.

The procedure is the same, but we’ll state it explicitly so that people can see it in action.

Let ${f(x)=2}$. We want to show that it is ${\displaystyle \lim_{x \rightarrow 1}f(x)=2}$.

{\begin{aligned} |f(x)-2| &< \delta \\ |2-2| &< \delta \\ 0 &< \delta \end{aligned}}

Which is trivially true for any value of ${\delta}$, hence ${\epsilon}$ can be any positive real number.

Let ${f(x)=2x+3}$. We want to show that it is ${\displaystyle \lim_{x \rightarrow 1}f(x)=5}$.

{\begin{aligned} |f(x)-5| &< \delta \\ |2x+3-5| &< \delta \\ |2x-2| &< \delta \\ 2|x-1| &< \delta \\ |x-1| &< \dfrac{\delta}{2} \end{aligned}}

With ${\epsilon=\delta/2}$ we satisfy the ${\epsilon}$ ${\delta}$ for limit.

As a final example let us look at the modified Dirichlet function that was introduced at this post.

$\displaystyle f(x) = \begin{cases} o \quad x \in \mathbb{Q}\\ x \quad x \in \mathbb{R}\setminus \mathbb{Q} \end{cases}$

At that post it was proved that for ${a \neq 0}$ ${\displaystyle\lim_{x \rightarrow a}f(x)}$ didn’t exist and it was promised that in a later date I’d show that ${\displaystyle\lim_{x \rightarrow 0}f(x)=0}$ using the epsilon delta condition.

Since we now know what the epsilon delta condition is and already have some experience with it will tackle this somewhat more abstruse problem.

{\begin{aligned} |f(x)-f(0)| &< \delta \\ |f(x)-0| &< \delta \end{aligned}}

Since ${f(x)=0}$ or ${f(x)=x}$ we have two cases to look at.

In the first case it is ${|0-0| < \delta}$ which is trivially valid, hence ${\epsilon}$ can be any real positive number.

In the second case it is ${|x-0| < \delta}$. Hence letting ${\epsilon=\delta}$ gets the job done.

Since we proved that ${\displaystyle\lim_{x \rightarrow 0}f(x)=0=f(0)}$ the conclusion is that the modified Dirichlet function that was presented is only continuous at ${x=0}$.

As was said previously, they don’t make local concepts more local than that.