## Real Analysis – Limits and Continuity IV

As an application of theorem 35 let us look into the functions ${f(x)=e^x}$ and ${g(x)=\log x}$.

Now ${f:\mathbb{R} \rightarrow \mathbb{R^+}}$ and is a strictly increasing function, and ${g:\mathbb{R^+} \rightarrow \mathbb{R}}$ also is a strictly increasing function.

By theorem 35 it is ${\displaystyle \lim_{x \rightarrow +\infty}\exp x = \mathrm{sup} [\mathbb{R^+}] = +\infty}$ and ${\displaystyle \lim_{x \rightarrow -\infty} \exp x= \mathrm{inf} [\mathbb{R^+}] = 0}$.

As for ${g(x)}$ it is ${\displaystyle \lim_{x \rightarrow +\infty} \log x=\sup [\mathbb{R}]=+\infty}$ and ${\displaystyle \lim_{x \rightarrow 0} \log x = \inf [\mathbb{R}]=-\infty}$.

 Definition 33 Let ${D \subset \mathbb{R}}$; ${f,g: D \rightarrow \mathbb{R}}$, and ${c \in D^\prime}$. Let us suppose that there exists ${h: D \rightarrow \mathbb{R}}$ such as ${f(x) = h(x)g(x) }$. If ${\displaystyle \lim_{x \rightarrow c} h(x)=1 }$ we say that ${f(x)}$ is asymptotically equal to ${g(x)}$ when ${x \rightarrow c}$ and write ${f(x) \sim g(x)\,\, (x \rightarrow c)}$. If ${\displaystyle \lim_{x \rightarrow c} h(x) = 0}$ we say that ${f(x)}$ is little-o of ${g(x)}$ when ${x \rightarrow c}$ and write ${ f(x) = o (g(x)) \,\, (x \rightarrow c)}$. If ${h(x)}$ is bounded in some neighborhood of ${c}$ we say that ${f(x)}$ is big-o of ${g(x)}$ when ${x \rightarrow c}$ and write ${f(x)=O(g(x)) \;(x \rightarrow c)}$.

If in the previous definition ${g(x)}$ doesn’t equal zero:

1. ${ f(x) \sim g(x) \Leftrightarrow \displaystyle \lim_{x \rightarrow c} \frac{f(x)}{g(x)} = 1}$.
2. ${ f(x) = o (g(x)) \,\, (x \rightarrow c) \Leftrightarrow \displaystyle \lim_{x \rightarrow c} \frac{f(x)}{g(x)} = 0}$.
3. ${ f(x) = O(g(x)) \,\, (x \rightarrow c) \Leftrightarrow \dfrac{f(x)}{g(x)} }$ is bounded in some neighborhood of ${c}$.

These notions work exactly as they worked for sequences and they give the same type of information about the behavior of the functions in question.

 Theorem 36 Let ${D \subset \mathbb{R}}$; ${f,g,f_0,g_0: D \rightarrow \mathbb{R}}$, and ${c \in D^\prime}$. Then: If ${f(x) \sim g(x) \,\, (x \rightarrow c)}$ and ${\displaystyle \lim_{x \rightarrow c}g(x) = a}$, then ${\displaystyle \lim_{x \rightarrow c} f(x) = a}$ If ${f(x) \sim f_0(x) \,\, (x \rightarrow c)}$ and ${g(x) \sim g_0(x) \,\, (x \rightarrow c)}$, then ${f(x)g(x) \sim f_0(x)g_0(x) \,\, (x \rightarrow c)}$ and ${f(x)/g(x) \sim f_0(x)/f_0(x) \,\, (x \rightarrow c)}$. Proof: Left as an exercise. $\Box$

As an example of the previous definitions we can say, with full generality, that for any polynomial function we can keep track of the term with the leading degree if we are interested in how it behaves for larger and larger values.

But on the other hand if we are interested on how the polynomial function behaves near the origin we have to keep track of the term with the smaller degree. To see that this is indeed so let us introduce the following example:

$\displaystyle f(x) = x^2+x$

Now ${x^2+x=(x+1)x}$. If we take ${h(x)=x+1}$ it is ${\displaystyle \lim_{x \rightarrow 0} h(x)=1}$ and so it is ${x^2+x=O(x) \,\, (x \rightarrow 0)}$.

Another example that has a lot of interest to us is:

$\displaystyle \sin x \sim x \,\, (x \rightarrow 0)$

We can see that it is so because of ${\displaystyle \lim_{x \rightarrow 0} \frac{\sin x}{x} = 1}$

— 6.6. Epsilon-delta condition —

And it is time for us to introduce the concept of limit using the ${ \epsilon - \delta }$ condition.

Once again we are walking into regions of greater and greater rigor at the expense of having to use more abstract concepts while we are doing it. Things are going to get a little harder for people that aren’t used to this types of reasoning but please bear with me and you’ll find it rewarding when you get used to it.

The point of the ${ \epsilon - \delta }$ condition is to avoid using fuzzy concepts near, input signals, output signals, or the somewhat weak definition of limit we been using so far.

 Theorem 37 (Heine’s Theorem) Let ${D \subset \mathbb{R}}$, ${f: D \rightarrow \mathbb{R}}$, ${c \in D^\prime}$ and ${a \in \overline{\mathbb{R}}}$. ${\displaystyle \lim_{x \rightarrow c} f(x) = a}$ if and only if $\displaystyle \forall \delta > 0 \, \exists \epsilon >0 : \; x \in V(c,\epsilon) \cap (D \setminus \left\lbrace c \right\rbrace ) \Rightarrow f(x) \in V(a, \delta)$ Proof: Omitted. $\Box$

In case you are wondering what that means the straightforward answer is that it means exactly what you’re idea of a function having a limit in a given point is (I’m assuming you have the right idea). It tell us that if a function indeed has limit ${a}$ in point ${c}$ then, if we restrict ourselves to points near ${c}$, the images of those points are all near ${a}$.

Once again I tell the reader to look at this as if it were a game played between two (slightly odd) people. One of them is choosing the ${\delta}$ and the the other is choosing the ${\varepsilon}$. But this game isn’t just about choosing. The first player gets to choose any ${\delta}$ he wants, but the second has to choose the right ${\varepsilon}$that makes the condition hold.

If he can prove that he has an ${\varepsilon}$ for every ${\delta}$ that the other player chooses than he succeeds in the game and the function does have limit ${a}$ at point ${c}$.

 Theorem 38 Let ${D \subset \mathbb{R}}$, ${f: D \rightarrow \mathbb{R}}$, and ${c \in D^\prime}$. If ${\displaystyle \lim_{x \rightarrow c} f(x)}$ exists and is finite, than there exists a neighborhood of ${c }$ where ${f(x)}$ is bounded. Proof: Let ${\displaystyle \lim_{x \rightarrow c} f(x) = a \in \mathbb{R}}$. By theorem 37 with ${\delta=1}$ there exists ${\varepsilon > 0}$ such as {\begin{aligned} x \in V(c,\varepsilon)\cap(D\setminus\left\lbrace c \right\rbrace ) &\Rightarrow f(x) \in V(a,1) \\ &\Rightarrow f(x) \in \left] a-1, a+1\right[ \end{aligned}} Thus ${x\in V(c,\varepsilon)\cap(D\setminus\left\lbrace c \right\rbrace)\Rightarrow a-1 < f(x) < a+1}$. So ${x \in V(c,\varepsilon) \cap D \Rightarrow f (x) \begin{cases} \leq \mathrm{max} \left\lbrace a+1,f(c)\right\rbrace \\ \geq \mathrm{max}\left\lbrace a+1,f(c)\right\rbrace \end{cases} }$ and ${f(x)}$ is bounded in ${V(c,\varepsilon)}$ $\Box$

If ${\displaystyle \lim_{x \rightarrow c} f(x)/g(x)}$ exists, then ${f(x)= O(g(x))\,\, (x \rightarrow c)}$ since in this case it is ${h(x)=f(x)/g(x)}$ and there exists some neighborhood of ${c}$ where ${h(x)}$ is bounded.

After this one may be interested in knowing how we can translate ${\displaystyle \lim_{x \rightarrow c^+} f(x) = a}$ to a ${\varepsilon - \delta}$ condition.

In this case we are considering ${f(x)}$ only in the set ${D_{c^+}}$ and so what we get is:

$\displaystyle \forall \delta > 0 \exists \varepsilon > 0: \, x \in V(c,\varepsilon)\cap D_{c^+} \Rightarrow f(x) \in V(a,\delta)$

 Theorem 39 Let ${D \subset \mathbb{R}}$, ${f:D \rightarrow \mathbb{R}}$, and ${c \in D^\prime}$. If ${\displaystyle \lim_{x \rightarrow c^-}f(x)=\lim_{x \rightarrow c^+}f(x)=a}$, then ${\displaystyle \lim_{x \rightarrow c}f(x)=a}$. Proof: Let ${\delta > 0}$. By the ${\varepsilon-\delta}$ condition it is: $\displaystyle \exists \varepsilon_1>0:x \in V(c,\varepsilon_1)\cap D_{c^+} \Rightarrow f(x) \in V(a,\delta)$ $\displaystyle \exists \varepsilon_2>0:x \in V(c,\varepsilon_2)\cap D_{c^-} \Rightarrow f(x) \in V(a,\delta)$ Thus by taking ${\varepsilon =\mathrm{min} \left\lbrace \varepsilon_1, \varepsilon_2 \right\rbrace }$ it follows ${x \in V(c,\varepsilon) \cap (D \setminus \left\lbrace c \right\rbrace ) \Rightarrow x \in V(c,\varepsilon) \cap D_{c^+}}$ or ${x \in V(c,\varepsilon) \cap D_{c^- }\Rightarrow f(x) \in V(a,\delta)}$ In conclusion: ${ \forall \delta > 0 \exists \varepsilon > 0: x \in V(c,\varepsilon)\cap (D\setminus \left\lbrace c \right\rbrace ) \Rightarrow f(x) \in V(a,\delta) }$ which is equivalent to saying that ${\displaystyle \lim_{x \rightarrow c} f(x)=a}$. $\Box$

 Definition 34 Let ${D \subset \mathbb{R}}$; ${f: D \rightarrow \mathbb{R}}$ and ${c \in D}$. We say that ${f(x)}$ is continuous in point ${c}$ if for all sequences ${x_n}$ of points in ${D}$, such as ${\lim x_n = c}$ it is ${\lim f(x_n)=f(c)}$. A function is said to be continuous if it is continuous in all points in ${D}$.

A few examples to clarify definition 34

1. $\displaystyle f(x)=|x| \quad \forall x \in \mathbb{R}$

Let ${c \in \mathbb{R}}$ and ${x_n}$ a sequence such as ${x \rightarrow c}$. Then ${f(x_n)=|x_n|}$ and ${\lim f(x_n) = \lim |x_n| = |c|}$. In conclusion ${f(x_n) \rightarrow f(c)}$ which is equivalent to saying that ${f}$ is continuous in ${c}$. Since ${c}$ can be any given point ${f(x)=|x|}$ is continuous in ${\mathbb{R}}$.

2. Let ${f(x)= \sin x}$ and ${x_n}$ a sequence such as ${x_n \rightarrow \theta}$. It is ${\lim \sin x= \sin \theta}$ and by the same reasoning ${\sin x}$ is also continuous.
3. In general if ${x_n \rightarrow c}$ it is ${\lim f(x_n)=f(c)=f(\lim x_n)}$. So for ${\exp (x)}$ it is ${\lim \exp (x_n)=\exp (\lim x_n)}$.

If ${x_n \rightarrow +\infty }$ it follows that ${\lim \exp(x_n)=+\infty }$ and for ${x_n \rightarrow -\infty}$ it follows that ${\lim \exp(x_n)=0}$.

Thus if we define ${\exp (+\infty)=+\infty}$ and ${\exp (-\infty)=0}$ it follows that it always is ${\lim \exp (x_n)=\exp (\lim x_n)}$.

4. Analogously we can define ${\log +\infty= +\infty}$ and ${\log 0 = -+\infty}$ and it always is ${\lim \log x_n = \log (\lim x_n)}$.
 Theorem 40 (Heine’s theorem for continuity) Let ${D \subset \mathbb{R}}$, ${f:D \rightarrow \mathbb{R}}$ and ${c \in D}$. ${f}$ is continuous in ${D}$ if and only if $\displaystyle \forall \delta>0 \,\,\exists \, \varepsilon > 0: \, x \in D \wedge |x-c| < \varepsilon \Rightarrow |f(x)-f(c)| < \delta$ Or written in terms of neighborhoods $\displaystyle \forall \delta>0 \,\,\exists \, \varepsilon > 0: \, x \in V(c,\varepsilon) \cap D \Rightarrow f(x) \in V(f(c),\delta)$ Proof: Omitted. $\Box$

As can be seen the ${\varepsilon - \delta}$ condition for continuity in point ${c}$ is very similar to the one for limit ${a}$ in point ${c}$.

To finish this post I’ll just state a theorem that sheds some light on the connections of these two concepts:

 Theorem 41 Let ${D \subset \mathbb{R}}$, ${f:D \rightarrow \mathbb{R}}$ and ${c \in D \cap D^\prime}$. Then ${f}$ it’s continuous in point ${c}$ if and only if ${\displaystyle \lim_{x \rightarrow c} f(x) = c}$. Proof: Omitted. $\Box$

So as this theorem shows the connection between continuity and limit is indeed a deep one, but we can look at the concept of limit as being an auxiliary tool to determine if a function is continuous or not and we should not confuse them.

In the next post I intend to write a little bit more about continuity but in the mean time a very good text about it can be found here