Real Analysis – Limits and Continuity II

Posted on March 1, 2024 by ateixeira

\displaystyle \lim_{x \rightarrow 0^+} \frac{1}{x}

In this case it is {D_{0^+} = \left] 0, +\infty\right[ } and { 0^+ \in D_{0^+} }.

If {x_n} is a sequence of points in {D_{0^+}} such as {x_n \rightarrow 0^+} it follows { \lim f(x_n) = \lim \dfrac{1}{x_n} = \dfrac{1}{0^+} = + \infty }.

After this simple example we’ll introduce a theorem that will state a somewhat obvious result.

In layman terms what it expresses is that if a function has a limit in a given point {c} than the one-sided limits have to be equal and equal to the limit of the function.

In a more kinematic way it tell us if we approach {c} by points of the domain from the right of {c}, or from the left of {c}, the images of those points have to converge to the same value.

Theorem 29

Let {D \subset \mathbb{R}}, {f: D \rightarrow \mathbb{R} }, {c \in D^\prime} and let us suppose that {\displaystyle \lim_{x \rightarrow c} f(x) = a}. Then, if {c \in D^\prime_{c^+}} it also is {\displaystyle \lim_{x \rightarrow c^+} f(x) = a}; and if {c \in D^\prime_{c^-}} it also is {\displaystyle \lim_{x \rightarrow c^-} f(x) = a}.

Proof:

Let {x_n} be a sequence of points in {D_{c^+} } such as {x_n \rightarrow c}. Since {x_n} is a sequence of points in {D \setminus \left\lbrace c \right\rbrace} (by our choice of {x_n}) and {\displaystyle \lim_{x \rightarrow c} f(x)=a } (by the hypothesis of the theorem) it follows from the definition of limit that { \lim f(x_n) = a }.

But this is just { \displaystyle \lim_{x \rightarrow c^+} f(x) = a} by definition 32.

The case { \displaystyle \lim_{x \rightarrow c^-} f(x) } is proved in the same way with the due modifications and is left as an exercise for the reader. \Box

\displaystyle \lim_{x \rightarrow 0} \dfrac{1}{x}

Using the previous theorem it is easy to see that this limit doesn’t exist. We already know that {\displaystyle\lim_{x \rightarrow 0^+}\dfrac{1}{x}=+\infty} and that {\displaystyle \lim_{x \rightarrow 0^-} \dfrac{1}{x} = - \infty }.

Since the limit from the right of {0} is different from the limit of the left of {0} we can conclude that this limit doesn’t exist.

Just in case the previous example has caused some doubts on the reader I’ll now try to explain in a more clear way the reasoning behind it.

Theorem 29 is an implication theorem. By that I mean a theorem that states a relationship between two propositions where the fact of the first proposition is true implies that the second proposition is also true.

So, we have proposition {P} and proposition {Q}. And an implication between those two propositions can be for example: “The validity of {Q} depends on the validity of {P}“.

What this means is that if {P} is a true statement than the validity of {Q} will follow.

And that if {Q} is a false statement than {P} will also be a false statement.

In mathematical notation: (where {\neg P} means not {P}) {P \Rightarrow Q \Leftrightarrow \neg Q \Rightarrow \neg P}.

Note that if we have {\neg P} we can’t conclude anything about the logical value of {Q} and that if we have {Q} we can’t conclude anything about the value of {P}.

An everyday situation may helps us here:

Imagine that you are waiting for and old friend from an uncle of yours called Pierre.

You have never known Pierre and the only thing that you know about him is that he only speaks French.

So a fellow comes to you and starts asking for directions in English. At that moment you can conclude that the fellow in question isn’t Pierre ({ \neg Q \Rightarrow \neg P }).

If by chance some fellow comes near you speaking French than you can’t conclude anything (remember that Pierre isn’t the only French speaking guy on Planet Earth).

In Theorem 29 we had { \displaystyle \lim_{x \rightarrow c} f(x)=a \Rightarrow\lim_{x \rightarrow c^+} f(x)=a \land \lim_{x \rightarrow c^-} f(x)=a }.

In this case {P} is { \displaystyle \lim_{x \rightarrow c} f(x) = a } and Q is { \lim_{x \rightarrow c^+} f(x) = a \land \lim_{x \rightarrow c^-} f(x) = a }.

So by showing that {\displaystyle \lim_{x \rightarrow 0^+} \dfrac{1}{x} \neq \lim_{x \rightarrow 0^-} \dfrac{1}{x}} we arrived at the conclusion that we have {\neg Q} and so {\neg P} has to follow.

In this case {\neg P} is just the statement that {\lim_{x \rightarrow 0}\dfrac{1}{x} = a} is a meaningless statement for any {a\in{\mathbb R}} and so {\lim_{x \rightarrow 0}\dfrac{1}{x}} doesn’t exist.

— 6.4. Limits of Functions and Inequalities —

We will now state a group of theorems that generalize what we already saw for sequences.

Theorem 30 (Limit of Inequalities) Let {D \subset \mathbb{R}}, {f,g : D \rightarrow \mathbb{R}}, {c \in D^\prime} and let us suppose that there exists {r > 0} such as {f(x) < g(x)\quad \forall x \in V(c,r) \cap (D\setminus \left\lbrace c \right\rbrace ) }.

If {\displaystyle \lim_{x \rightarrow c} f(x)} and {\displaystyle \lim_{x \rightarrow c} g(x)} exist it is {\displaystyle \lim_{x \rightarrow c} f(x) \leq \lim_{x \rightarrow c} g(x)}

Proof:

Let {x_n} be a sequence of points in {D \setminus \left\lbrace c \right\rbrace } such as {x_n \rightarrow c}. By definition 18 {\exists k \in \mathbb{N}:\quad n \geq k \Rightarrow x_n \in V(c,r) \Rightarrow x_n \in V(c,r) \cap D\setminus \left\lbrace c \right\rbrace }.

Since {x \in V(c,r) \cap D \setminus \left\lbrace c \right\rbrace \Rightarrow f(x) \leq g(x)}.

Thus {n \geq k} implies that {f(x_n) \leq g(x_n)}.

By Theorem 14 we know that it is {\displaystyle \lim f(x_n) \leq \lim g(x_n)}.

Since {\displaystyle \lim_{x \rightarrow c} f(x) = f(x_n)} and {\displaystyle \lim_{x \rightarrow c} g(x) = g(x_n)} it follows {\displaystyle \lim_{x \rightarrow c} f(x) \leq \lim_{x \rightarrow c} g(x)} \Box

Corollary 31

Let {D \subset \mathbb{R}}, {f: D \rightarrow \mathbb{R} }, {c \in D^\prime} and {a \in \mathbb{R}}.

If there exists {r > 0} such as {f(x) \leq a} ({f(x) \geq a}) { \forall x \in V(c,r) \cap D \setminus \left\lbrace c \right\rbrace} and if {\displaystyle \lim_{x \rightarrow c} f(x)} exist. It is { \displaystyle \lim_{x \rightarrow c} f(x) \leq a} ({ \displaystyle \lim_{x \rightarrow c} f(x) \geq a }).

Proof: Take {g(x)=a} in the previous theorem. \Box

About ateixeira

Experienced professional with a track record on strategic roles in several industries in Angola and other emerging markets. Right now I am getting back to Physics and this blog is a way to keep track of my progress and struggles
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