## Real Analysis – Limits and Continuity III

The first thing I want to say is that the concept of limit is a local concept.

In mathematical lingo what this means is that for a function to have a limit in a given point, ${\displaystyle \lim_{x \rightarrow c} f(x) = a}$, it doesn’t matter how the function behaves when we are far away from the point in question, what matters is just how the function behaves in the vicinity of the point.

This all very good for common day to day knowledge but it is not good enough for Mathematics.

So, with the concept of limit what we are doing is formalizing what we mean with the expressions far away and vicinity.

For an example let me introduce the function

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

This function’s not a very sophisticated but it’s good enough for what I’m trying to convey.

First of all let us plot this function to see what it looks like.

Where we have drawn the ${x \in \mathbb{Q}}$ case in blue and the ${x \in \mathbb{R}\setminus \mathbb{Q}}$ case in red.

It is easy to see that for all ${c}$ different than ${0}$ the function has no limit.

For ${c \neq 0}$ ${ \displaystyle \lim_{x \in \mathbb{Q} \rightarrow c} f(x) = 0 }$ and ${ \displaystyle\lim_{x \in \mathbb{R}\setminus \mathbb{Q}\rightarrow c} f(x) = c }$. Thus ${ \displaystyle \lim_{x \in \mathbb{Q} \rightarrow c} f(x) \neq \lim_{x \in \mathbb{R}\setminus \mathbb{Q}\rightarrow c} f(x)}$, and we can conclude that this limit doesn’t exist.

For ${c=0}$ it is possible to prove (I’ll do that when the concept of limit is formalized using an ${ \epsilon-\delta }$ condition) that ${ \displaystyle \lim_{x \rightarrow 0} f(x) = 0}$.

They don’t make concepts more local than this! This function only has a limit at point ${0}$.

In an intuitive way we can understand this result like this: we can think that the concept of limit is a measure of how good behaved a function is.

Since this function is always jumping from point to point as we move from rational numbers to irrational numbers we can say that it isn’t a well behaved one.

The former statement is true almost everywhere in the domain of the function. The only point where it breaks down is at point ${0}$.

This is so because even though the function is a badly behaved one it misbehaves less and less while ${x \rightarrow 0}$.

 Theorem 32 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 ${f(x) \leq g(x)\quad \forall x \in V(c,r) \cap \left( D \setminus \left\lbrace c\right\rbrace \right)}$. Then, if ${\displaystyle \lim_{x \rightarrow c} f(x)= +\infty }$ it also is ${\displaystyle \lim_{x \rightarrow c} f(x)= +\infty }$. And if ${\displaystyle \lim_{x \rightarrow c} g(x)= -\infty }$ it also is ${\displaystyle \lim_{x \rightarrow c} f(x)= -\infty }$ Proof: Omitted. $\Box$

The previous theorem states a very straightforward fact, but, as always, what matters is that this result can be proven. In more prosaic terms this theorem expresses the conditions that need to fulfilled for us to know the limits of some functions just by knowing the limit of another one.

It may well be the case that one limit may be very easy to calculate while the other is not.

If we can establish an order relationship and calculate one of the limits it is possible for us to conclude something about the limit of the other function. In Theorem 32 we where particularly interested in the cases when the limit is ${ \pm \infty}$ but we already seen in Theorem 30 that limit weakens order relationships.

In this case if it is ${f(x)\leq g(x)}$, for some neighbourhood around a point ${c}$, then we know that ${\displaystyle \lim_{x \rightarrow c} f(x)\leq \lim_{x \rightarrow c} g(x)}$ also. Now, if ${\displaystyle\lim_{x \rightarrow c}f(x)=+\infty}$ ${g(x)}$ has no choice but to go to positive infinity as we move closer to ${c}$ since it has to be larger than ${f(x)}$.

In the case ${\displaystyle \lim_{x \rightarrow c} g(x) = -\infty}$ a similar reasoning applies.

${f(x)}$ is smaller than ${g(x)}$ and if ${g(x)}$ gets to smaller and smaller values as we approach ${c}$ than ${f(x)}$ also has to get smaller and smaller values.

 Theorem 33 (Squeezed function theorem) 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 ${g(x) \leq f(x) \leq h(x)\quad \forall x \in V(c,r) \cap D \setminus \left\lbrace c \right\rbrace }$. If ${\displaystyle \lim_{x \rightarrow c} g(x) = \lim_{x \rightarrow c} h(x) = a }$ it also is ${\displaystyle \lim_{x \rightarrow c} f(x) = a}$. Proof: Omitted. $\Box$

This theorem continues the trend of computing limits of functions without computing them!

In here if we can box the function in a neighbourhood of a point by two functions, and if we compute the limits of the boxing functions and come to the conclusion that they are equal we are able to know that the boxed function has the same limit.

As an example let us see the limit:

$\displaystyle \lim_{x \rightarrow +\infty} \frac{\sin x}{x}$

It is ${-1 \leq \sin x \leq 1 \quad \forall x \in \mathbb{R}}$. Thus ${\displaystyle -\frac{1}{x} \leq \frac{\sin x}{x} \leq \frac{1}{x} \quad \forall x > 0}$.

Since ${\displaystyle \lim_{x \rightarrow +\infty}-\frac{1}{x}=\lim_{x \rightarrow +\infty}\frac{1}{x}= 0}$ it also is ${\displaystyle \lim_{x \rightarrow +\infty}-\frac{\sin x}{x}=0}$.

As a second example let us now look into:

$\displaystyle \lim_{x \rightarrow 0} \frac{\sin x}{x}$

Since ${\displaystyle \cos x < \frac{\sin x}{x} < 1\quad \forall x \in \left] -\frac{\pi}{2},0 \right[ \cup \left] 0,\frac{\pi}{2}\right[ }$

It is ${\displaystyle \lim_{x \rightarrow 0}1=1}$ and ${\displaystyle \lim_{x \rightarrow 0} \cos x = \cos 0 = 1}$. Thus it also is ${\displaystyle \lim_{x \rightarrow 0} \frac{\sin x}{x}=1}$

— 6.5. Algebraic Properties of the Limit of Functions —

Just like we did for sequences we’ll derive the algebraic rules that allows to manipulate the limits of some more complex expressions.

 Theorem 34 (Algebraic properties of limits) Let ${D \subset \mathbb{R}}$; ${f,g:D \rightarrow \mathbb{R}}$ and ${c \in D^\prime}$. Then: ${\displaystyle \lim_{x \rightarrow c} f(x)=a \Rightarrow \lim_{x \rightarrow c} |f(x)|=|a|}$ ${\displaystyle \lim_{x \rightarrow c} f(x)=a}$ and ${\displaystyle \lim_{x \rightarrow c} g(x)=b}$, then ${\displaystyle \lim_{x \rightarrow c} \left( f(x)+g(x)\right) = a+b}$ If ${\displaystyle \lim_{x \rightarrow c} f(x) = +\infty }$ and ${g}$ bounded below, than ${\displaystyle \lim_{x \rightarrow c} (f(x)+g(x))= +\infty}$ If ${\displaystyle \lim_{x \rightarrow c} f(x) = -\infty }$ and ${g}$ bounded above, than ${\displaystyle \lim_{x \rightarrow c} (f(x)+g(x))= -\infty}$ If ${\displaystyle \lim_{x \rightarrow c} f(x) = 0 }$ and ${g}$ bounded, than ${\displaystyle \lim_{x \rightarrow c} (f(x)g(x))= 0}$ If ${\displaystyle \lim_{x \rightarrow c} f(x) = a }$ and ${\displaystyle \lim_{x \rightarrow c} g(x) = b}$, than ${\displaystyle \lim_{x \rightarrow c} (f(x)g(x))= ab}$ If ${\displaystyle \lim_{x \rightarrow c} f(x) = +\infty }$ and ${\displaystyle \lim_{x \rightarrow c} g(x) = a \neq 0}$, than ${\displaystyle \lim_{x \rightarrow c} |f(x)g(x)|= +\infty}$ If ${\displaystyle \lim_{x \rightarrow c} f(x) = a \neq 0 }$, than ${\displaystyle \lim_{x \rightarrow c} 1/f(x)= 1/a}$ If ${\displaystyle \lim_{x \rightarrow c} f(x) = +\infty }$, than ${\displaystyle \lim_{x \rightarrow c} 1/f(x)= 0}$ If ${\displaystyle \lim_{x \rightarrow c} f(x) = 0 }$, than ${\displaystyle \lim_{x \rightarrow c} 1/|f(x)|= +\infty}$ Proof: We’ll only prove the second one since the reasoning is mostly the same for all propositions. Let ${x_n}$ be a sequence in ${D \setminus \left\lbrace c \right\rbrace }$ such as ${x_n \rightarrow c}$. Then ${f(x_n) \rightarrow a}$ and ${g(x_n) \rightarrow b}$. And from what we already saw for sequences it is ${f(x_n)+g(x_n) \rightarrow a+b}$. By definition of limit it is ${\displaystyle \lim_{x \rightarrow c} (f(x)+g(x)) = a + b}$. $\Box$
 Theorem 35 (Monotone function Theorem) Let ${D \subset \mathbb{R}}$; ${f: D \rightarrow \mathbb{R}}$, ${ \alpha = \inf D}$ and ${ \beta = \sup D}$. Then: If ${ \alpha \in D^\prime }$, ${\displaystyle \lim_{x \rightarrow \alpha} f(x)}$ exists and it is: ${\displaystyle \lim_{x \rightarrow \alpha} f(x) = \mathrm{inf}f \left[ D_{\alpha^+} \right] }$ if ${f}$ is increasing. ${\displaystyle \lim_{x \rightarrow \alpha} f(x) = \mathrm{sup}f \left[ D_{\alpha^+} \right] }$ if ${f}$ is decreasing. If ${ \beta \in D^\prime }$, ${\displaystyle \lim_{x \rightarrow \beta} f(x)}$ exists and it is:${\displaystyle \lim_{x \rightarrow \alpha} f(x) = \mathrm{sup}f \left[ D_{\beta^-} \right] }$ if ${f}$ is increasing.${\displaystyle \lim_{x \rightarrow \alpha} f(x) = \mathrm{inf}f \left[ D_{\beta^-} \right] }$ if ${f}$ is decreasing. Proof: A formal proof of this theorem won’t be given but I’ll provided a plot of a function to help us visualize this theorem. As an example of an increasing function ${f(x) = \sin x \quad \forall x \in \left] -\pi/2, \pi/2\right[ }$ In this case it is ${ \alpha = -\pi/2 }$ and ${ \beta = \pi/2 }$; ${\displaystyle \lim_{x \rightarrow -\pi/2} \sin x = \sin(-\pi/2)= -1}$.${ D_{\alpha^+}}$ represents ${D\cap \left] \alpha, +\infty \right[}$ So that ${f \left[ D_{\alpha^+} \right] }$ represents the image of ${f}$ by ${ D \cap \left] \alpha, +\infty \right[ }$. That is to say that ${f \left[ D_{\alpha^+} \right] = \left] -1, 1 \right[ }$ and ${ \mathrm{inf}\left] -1, 1 \right[=-1 }$ as we already seen when calculating the limit. In a similar way we can also check that it indeed is ${\displaystyle \lim_{x \rightarrow \pi/2} \sin x = \sin(\pi/2)= f \left[ D_{\beta^-} \right]}$ For the decreasing function, ${f(x)= \cos x \quad \forall x \in ]0,\pi[}$, both steps are to be done by the reader. $\Box$