Usuario:Greenbreen/incubadora

Introduction

Now that we have reviewed functions, we arrive at a central idea of calculus, the concept of a limit. It is preferable to begin with a function $f(x)=x^{2}$. We know that $f(2)=4$. However, we can be a little more ingenious and create a "hole" at $x=2$. We can do this by subtly changing the function as follows:

$f(x)={\frac {x^{2}(x-2)}{x-2}}$

This last function is equal to $x^{2}$ everywhere except at $x=2$, where it does not exist. Now, a curious fact is that as $x$ gets closer to $2$, $f(x)$ gets closer to $4$. This is a useful fact, and we can express it in symbols as

$\qquad \lim _{x\to 2}f(x)=4$

Note that it doesn't matter what $f(x)$ is at $x=2$; in this case we have left it undefined, but it could have been $2$ or $15$ or $10000000$. It doesn't matter--the idea of a limit is that you are able to talk about how a function behaves when it is made closer and closer to a value without talking about how it behaves at the value. Now using variables we are able to say that $L$ is the limit of a function $f(x)$ when $x$ approaches $c$ if $f(x)\approx L$ when $x\approx c$.

We say that the limit as $x$ approaches $c$ of $f(x)$ is $L$, if $L$ exists as a finite number. And we express it algebraically as follows

$\lim _{x\to c}f(x)=L$

Intuitively, the limit $L$ is simply the number to which $f(x)$ becomes closer and closer as $x$ approaches $c$, but $f(c)$ doesn't need to be defined.

This idea of talking about a function when it approaches something was an important discovery, because it allows one to talk about things that before weren't possible. For example, let us consider the function $1/x$. When $x$ is made very large, $1/x$ is made very small, closer and closer to zero, the larger $x$ is made. Without limits it is very difficult to talk about this fact, because $1/x$ never really reaches zero. But the language of limits exists precisely for talking about the behavior of a function when it is close to something, without worrying that it never gets there. So we can say

$\qquad \lim _{x\to \infty }{\frac {1}{x}}=0$

The application of calculus to instantaneous velocity

To see the power of limits, let's go back to the example of the moving car that we mentioned in the introduction. Suppose that we have a car whose position is linear with time. We want to find the velocity. This is easy to do with algebra, simply take the slope of the straight-line distance verses time of the car, and this is our velocity.

But unfortunately (or perhaps fortunately if you are a professor of calculus), things in the real world don't always travel in nice straight lines. Cars accelerate, decelerate, and generally behave in ways that make it difficult to calculate their velocities. (figure 2).

Now what we really want is to find the velocity at a given moment. (figure 3). The problem is that to find the velocity we need two points, while at any given time we only have one point. We can, of course, always find the average velocity of the car, given two points in time, but we want to find the velocity of the car at a precise moment.

This is where the basic trick of differential calculus enters. We choose a pair of points on our graph of position vs. time, one of which is at the time in which we want to find the instantaneous velocity and the other at any other place, and then we draw a line between them. We begin to bring this last point closer and closer to the first and to draw successive lines between them. As the points are made closer, the slopes of the line between them approach the instantaneous velocity.

Formal definition of the limit

The formal definition of the limit has a tradition of being somewhat complicated for students who see it for the first time. We will present it first and then we will see in detail what it so succinctly says.

We say that the limit of $f(x)$ as $x$ tends to $c$ is equal to $L$ or

$\lim _{x\to c}f(x)=L$

if for every $\epsilon >0$ there exists a $\delta >0$ such that, for all $x$, if $0<\left|x-c\right|<\delta$, then $\left|f(x)-L\right|<\epsilon$

Let's look at the rules for absolute values as they concern the previous expression. On the real line, the absolute value of a number is the distance between the number (be it positive or negative) and the origin. $-3$ is $3$ units of distance from zero, therefore $\left|-3\right|=3$. Similarly, the absolute value of a difference is the distance between the two numbers involved in it. Note that $\epsilon$ and $\delta$ in the previous expression define the distances between $f(x)$ and $L$ and between $c$ and $x$.

In other words, once the distance between $x$ and $c$ is chosen so that it is less than $\delta$ but greater than zero (because $x$ is close to $c$ but never reaches it), we can guarantee that the distance between $f(x)$ and $L$ is less than $\epsilon$ independently of the $\epsilon$ chosen (I DON'T THINK THIS STATEMENT IS RIGHT--I'LL HAVE TO ANALYZE IT LATER.)

One-sided limits

Observe that on the real line we are able to approach a particular value of $x$ either from larger values (from the right) or smaller values (from the left). Because there are functions that do not exhibit the same behavior from the left and right of a given value, the concept of one-sided limits may help to elucidate this behavior. Let us consider for example the function that takes the floor of $x$ (frequently used to explain the idea of one-sided limits), denoted by $\left\lfloor x\right\rfloor$ and defined to be the greatest integer less than or equal to $x$, so that

$\left\lfloor x\right\rfloor \leq x<\left\lfloor x\right\rfloor +1$

Its graph is a series of horizontal segments at distinct heights. Now let us consider as an example the particular integer value $1$. When we approach $1$ from the left we see that the successive values of $\left\lfloor x\right\rfloor$ are equal to zero, and as we approach $1$ from the right the values are one. The idea of successively approaching a value introduces the notion of a one-sided limit. In what follows we introduce the formal notions of one-sided limits from the left and right.

Limits from the right

The limit of $f(x)$ as $x$ approaches $a$ from the right is equal to $L$ if $\forall x>0$, there exists a $\delta >0$ such that if $0<x-a<\delta$, then $\left|f(x)-L\right|<\epsilon$. This is denoted as

$\lim _{x\to a^{+}}f(x)=L$

Limits from the left

The limit of $f(x)$ as $x$ approaches $a$ from the left is equal to $L$ if $\forall x>0$, there exists a $\delta >0$ such that if $0<a-x<\delta$, then $\left|f(x)-L\right|<\epsilon$. This is denoted as

$\lim _{x\to a^{-}}f(x)=L$

THEOREM The limit exists if and only if the two one-sided limits (from the right and from the left) both exist and agree with one another. Note that it is also valid if we consider that the limit is $+\infty$ or $-\infty$ instead of $1$. (*I'M NOT SURE WHAT THEY MEAN HERE*).

Fundamental theorems about limits

Let $f$ and $g$ be functions with limits at $c$, $n$ be an integer, and $k$ be a constant. Then we have the following:

• The limit of a constant is the constant:

$\lim _{x\to c}k=k$

• $\lim _{x\to c}x=c$

• The limit of a constant times a function is equal to the constant times the limit of the function:

$\lim _{x\to c}kf(x)=k\lim _{x\to c}f(x)$

• The limit of a sum is equal to the sum of the limits:

$\lim _{x\to c}[f(x)\pm g(x)]=\lim _{x\to c}f(x)\pm \lim _{x\to c}g(x)$

• The limit of a product is equal to the product of the limits:

$\lim _{x\to c}[f(x)\cdot g(x)]=\lim _{x\to c}f(x)\cdot \lim _{x\to c}g(x)$

• The limit of a quotient is the quotient of the limit provided that the limit of the denominator does not evaluate to zero:

$\lim _{x\to c}{\frac {f(x)}{g(x)}}={\frac {\lim _{x\to c}f(x)}{\lim _{x\to c}g(x)}}$ ,provided that $\lim _{x\to c}g(x)\neq 0$

• The limit of the power of a function is equal to the $n$th power of the limit of the function:

$\lim _{x\to c}[f(x)]^{n}=[\lim _{x\to c}f(x)]^{n}$

• The limit of the $n$th radical of a function is equal to the $n$th radical of the limit of the function:

$\lim _{x\to c}{\sqrt[{n}]{f(x)}}={\sqrt[{n}]{\lim _{x\to c}f(x)}}$

Proofs

Theorem of the sum of limits

We need to show that:

(1) $\lim _{x\to c}(f(x)+g(x))=\,L_{1}+L_{2}\,$

given that:

$\lim _{x\to c}f(x)=\,L_{1}\,;\lim _{x\to c}g(x)=\,L_{2}\,$

According to the definition of the limit, we have to have a $\delta$ for every $\epsilon$ such that:

(2) $0<|x-c|<\delta \longrightarrow |(f(x)+g(x))-(L_{1}+L_{2})|<\epsilon$

Then, in taking the limits of $f(x)$ and $g(x)$ as $x$ approaches $c$, we take an ${\frac {\epsilon }{2}}$ such that:

$0<|x-c|<\delta _{1}\longrightarrow |f(x)-L_{1}|<{\frac {\epsilon }{2}}$

and

$0<|x-c|<\delta _{2}\longrightarrow |g(x)-L_{2}|<{\frac {\epsilon }{2}}$

If we now take $\delta$ to be the lesser of $\delta _{1}$ and $\delta _{2}$ (i.e. $\delta \leq \delta _{1}$; $\delta \leq \delta _{2}$), then:

(3) $0<|x-c|<\delta \longrightarrow |f(x)-L_{1}|<{\frac {\epsilon }{2}}$

and

(4) $0<|x-c|<\delta \longrightarrow |g(x)-L_{2}|<{\frac {\epsilon }{2}}$

Note that:

$|(f(x)+g(x))-(L_{1}+L_{2})|=|(f(x)-L_{1})+(g(x)-L_{2})|\,$

Using the Triangle Inequality:

$|(f(x)+g(x))-(L_{1}+L_{2})|\leq |(f(x)-L_{1})|+|(g(x)-L_{2})|\,$

Replacing for (3)

y (4)

$|(f(x)+g(x))-(L_{1}+L_{2})|<{\frac {\epsilon }{2}}+{\frac {\epsilon }{2}},$

That is to say:

$|(f(x)+g(x))-(L_{1}+L_{2})|<\epsilon$

which proves (2)

, i.e., there is a $\delta$ for each $epsilon$ such that:

$0<|x-c|<\delta \longrightarrow |(f(x)+g(x))-(L_{1}+L_{2})|<\epsilon$

that according to (1)

verifies that:

$\lim _{x\to c}(f(x)+g(x))=\,L_{1}+L_{2}\,$

i.e.:

$\lim _{x\to c}(f(x)+g(x))=\,\lim _{x\to c}f(x)+\lim _{x\to c}g(x)\,$

The limit of the sum of two functions is equal to the sum of their limits.

The Sandwich Theorem

The sandwich theorem is useful when we want to find the limit of a function that is "confined" between two functions that we already know. If $f(x)\leq g(x)\leq h(x)$ y $\lim _{x\to c}f(x)=\lim _{x\to c}h(x)=L$, then $\lim _{x\to c}g(x)=L$

Limits of trigonometric functions

Demonstration of the limit of sin(x)/x when x approaches 0

Consider that, for small $x$, $\sin(x)\leq x\leq \tan(x)$

If we divide all the members by $\sin(x)$ what remains is ${\frac {\sin(x)}{\sin(x)}}\leq {\frac {x}{\sin(x)}}\leq {\frac {\tan(x)}{\sin(x)}}$

But ${\frac {\sin(x)}{\sin(x)}}=1$ and ${\frac {\tan(x)}{\sin(x)}}={\frac {\frac {\sin(x)}{\cos(x)}}{\sin(x)}}={\frac {1}{\cos(x)}}$

Inverting each member leaves us with this totally indeterminate expression $1\geq {\frac {\sin(x)}{x}}\geq \cos(x)$

If we find the limit as $x$ approaches 0 for each member we are left with $\lim _{x\to 0}1\geq \lim _{x\to 0}{\frac {\sin(x)}{x}}\geq \lim _{x\to 0}\cos(x)$

And since $\lim _{x\to 0}1=1$ and $\lim _{x\to 0}\cos(x)=1$, by the Sandwich Theorem we are left with $\lim _{x\to 0}{\frac {\sin(x)}{x}}=1$

Infinite limits

Consider the value of a function only at the numbers 0,1,2,3,4,5....evaluating the function at increasing or decreasing values in this manner. The function may approach a limit as the numbers increase or decrease toward infinity without ever touching the limit.

Definition 1

Let $f$ be a function that is defined on all numbers in some open interval $(a,+\infty )$. The limit of $f(x)$ when $x$ increases without limit is $L$, which is written as:

$\lim _{x\to +\infty }f(x)=L$

Definition 2

Let $f$ be a function that is defined on all numbers in some open interval $(-\infty ,a)$. The limit of $f(x)$ when $x$ decreases without limit is $L$, which is written as:

$\lim _{x\to -\infty }f(x)=L$

Theorem

Let $n$ be any positive integer. Then,

$\lim _{x\to +\infty }{\frac {1}{x^{n}}}=0$

$\lim _{x\to -\infty }{\frac {1}{x^{n}}}=0$

Limits to infinity

The challenge now is to calculate the value, when it exists and is finite, to which advanced terms of a sequence approach. We will use an enlightening example to introduce this. Consider the following sequence:

$a_{n}={\frac {1}{n}}$

If we write down some terms, we will quickly see the real value to which those terms approach:

$a_{1}=1,a_{2}={\frac {1}{2}}=0.5,a_{3}={\frac {1}{3}}=0.33333...,a_{4}={\frac {1}{4}}=0.25,a_{5}={\frac {1}{5}}=0.2,...$

As we can see, the terms are becoming smaller with each iteration so that, if the sequence had a final term, we would expect it to be 0. This gives us an intuitive idea of the meaning of the limit of a sequence as it tends to infinity; that is, to determine in some way to what value the terms of the sequence approach as we advance through them.

With the previous sequence, we can write $\lim _{n\to \infty }a_{n}=0$, and in fact, we can think of it intuitively as:

$\lim _{n\to \infty }a_{n}={\frac {1}{\infty }}=0$

Proving this equality is complicated, but we can illustrate it with the following example:

Suppose that we have a loaf of bread with which we are going to feed all the people in China. The question is how much bread does each person get.

Considering the number of Chinese people there are, we will have to give each a microscopic fraction of bread, with the portions almost molecular so that for all intents and purposes we can say that each person receives no food. The idea is that to divide one quantity by another that is immensely larger, the result is immensely small, so that to divide a number by infinity, which would be the largest of all numbers, would give us the least of all numbers, which is zero.

It is of significance for the calculation of limits to infinity to keep in mind some equalities involving infinity. We should always keep in mind that infinity is not a number, but a concept. We refer to infinity as that which is boundless, but through a harmless abuse of language we can use it for operations:

${\frac {k}{\infty }}=0$, $\forall k\neq 0,\infty$

${\frac {k}{0}}=\infty$, $\forall k\neq 0,\infty$

$\infty \pm k=\infty$, $\forall k\in \mathbb {R}$

• • Such statements really aren't harmless, e.g. as with the recently made comment, ${\frac {k}{0}}$ doesn't exist because the two-sided "limits" don't agree**

Indeterminates

There are limits that, on evaluating them, yield one of the following expressions:

$\infty -\infty ,\quad {\frac {\infty }{\infty }},\quad \infty \cdot 0,\quad {\frac {0}{0}},\quad \infty ^{0},\quad 1^{\infty },\quad 0^{0}$

These expressions are called indeterminates, since, at first glance, it is not clear what the limit is (if it exists). For example, in the second of these expressions, the limit could have the value 0, 1 or infinity. In some cases, simplifying the expression or obtaining expressions that are equivalent to the original through rationalization or factorization can resolve the indeterminacy and calculate the limit. In other cases, one is required to use more powerful tools such as inequalities or L'Hôpital's rule.

An example of indeterminacy of the type ${\frac {0}{0}}$ is given in these three cases, and in each case (after simplification), one obtains a different limit:

$\lim _{t\rightarrow 0}{\frac {t}{t^{2}}}={\frac {0}{0}}\quad {\xrightarrow[{\mathrm {simplificando} }]{}}\quad \lim _{t\rightarrow 0}{\frac {1}{t}}=\infty$

$\lim _{t\rightarrow 0}{\frac {t}{t}}={\frac {0}{0}}\quad {\xrightarrow[{\mathrm {simplificando} }]{}}\quad \lim _{t\rightarrow 0}1=1$

$\lim _{t\rightarrow 0}{\frac {t^{2}}{t}}={\frac {0}{0}}\quad {\xrightarrow[{\mathrm {simplificando} }]{}}\quad \lim _{t\rightarrow 0}{t}=0$

Asymptotes

Asymptotes are lines that a function approaches in the limit as $x$ or $f(x)$ tends to infinity.

Using the notation of limits to describe asymptotes

Now consider the function

$g(x)={\frac {1}{x}}.$

What is the limit as $x$ approaches zero? The value of $g(0)$ doesn't exist since

$\qquad g(0)={\frac {1}{0}}$

is not defined.

But it is intuitively clear that we can make the function $g$ as large as we want, choosing an $x$ sufficiently small. For example, to make $x$ equal to a million, choose $x$ to be $10^{-6}$. In this case we say that we can make $g(x)$ arbitrarily large (as large as we want) taking a $x$ to be sufficiently near to, but not equal to, zero, and we express this algebraically as follows:

$\lim _{x\to 0^{+}}g(x)=\lim _{x\to 0^{+}}{\frac {1}{x}}=\infty$

Note that we have used the right-sided limit notation, since the limit doesn't exist at $x=0$ (given that the limits from the left and right are distinct).

Similarly we can consider that as $x$ becomes larger and larger $g(x)$ tends to zero but never reaches it. This allows us to introduce vertical and horizontal asymptotic notation.

Vertical asymptotes

A vertical asymptote occurs at $x=c$, for a function $f(x)$ provided that:

1. $\lim _{x\to -3^{-}}f(x)=\pm \infty$
2. $\lim _{x\to -3^{+}}f(x)=\pm \infty$

It's worth emphasizing that at $c$ the function can be positive infinity or negative infinity (either of them, but not both) in order to have a vertical asymptote, and that this can be accomplished when we approach $c$ from the left (first case) or the right (second case).

(***Lose the -3***)
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