Theory of Limits

Idea of the Limit:

A number $L$ is the $limit$ of $F(x)$ as $x$ approaches $a$ , if the number $F(x)$ can be made as close to $L$ as we choose $x$ sufficiently near but not equal to $a$ .

In this case we get $F(x)$ closer and closer to $L$ when $x$ gets closer and closer to $a$ .

Many people make the mistake of thinking that $x=a$ . This is not true when it comes to the definition of limits.

The number $L$ is the $limit$ of $F(x)$ as $x$ approaches $a$ if, given any number $\varepsilon>0$ , there is a number $\delta>0$ such:

$|F(x)-L|<\varepsilon$

For all $x$ such that:

$0<|x-a|<\delta$

Example:

Evaluate

$\lim_{x\to 2}x^{3}$

If we dress a table of values of $x$ closer and closer to 2, from left or right, we can see that $x^{3}$ becomes closer and closer to $8$

It is always a good idea to investigate values close to $x$ and come to the conclusion of what the limit is.

As this will serve as a refresher, we are going to state the limit Laws.

Limit Laws

Limit of a constant C:

$\lim_{x\to a}C=C$

Addition, product and quotient Laws:

Let:

$\lim_{x\to a}F(x)=L$

and

$\lim_{x\to a}G(x)=K$

We can say:

Addition

$\lim_{x\to a}[F(x) \pm G(x)]=L \pm K$

Product

$\lim_{x\to a}[F(x)G(x)]=LK$

Quotient

$\lim_{x\to a}\frac{F(x)}{G(x)}=\frac{L}{K}$

With $K\neq 0$

If $n$ is a positive integer and $a>0$ the can have the following Law.

$\lim_{x\to a}\sqrt[n]{x}=\sqrt[n]{a}$

Substitution:

Suppose that:

$\lim_{x \to a}g(x)=L$

and

$\lim_{x \to L}f(x)=f(L)$

We can write:

$\lim_{x \to a}f(g(x))=f(\lim_{x \to a}g(x))=f(L)$

EXAMPLE

Prove:

$\lim_{x \to 5}(x^{2}-8)=17$

Solution:

Using the Definition above:

Given $\varepsilon>0$ , we have to find $\delta$ such

$0<|x-5|<\delta$ implies $|(x^{2}-8)-17|<\varepsilon$

We know:

$|(x^{2}-8)-17|=|x^{2}-25|=|x+5| \cdot |x-5|$

In our situation, if we make $|x-5|$ sufficiently small, $|x+5|$ cannot be too large.

If $|x-5|<1$

$|x+5|=|(x-5)+10|\leq |x-5|+10<11$

This means

$0<|x-5|<\delta$ implies $|(x^{2}-8)-17|<11 \cdot |x-5|$

Now if $\delta$ is the minimum of the two numbers $1$ and $\frac{\varepsilon}{11}$

$0<|x-5|<\delta$ implies $|(x^{2}-8)-17|<11 \cdot \frac{\varepsilon}{11}$

Or simply:

$0<|x-5|<\delta$ implies $|(x^{2}-8)-17|<\varepsilon$

One-sided limits:

One of the examples of left and right hand limits is:

$f(x)=\frac{x}{|x|}$

This function is simply $1$ for $x>0$ and $-1$ if $x<0$

The right-hand limit and the left-hand limit do not agree in this situation.

We conclude that the limit does not exist.

The Right-Hand Limit of a Function

Suppose that $f$ is defined on an open interval (a,c).

L is the right-hand limit of $f(x)$ as $x$ approaches $a$ , and we have:

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

If $f(x)$ can be made as close to $L$ by chossing a point $x$ in $(c,a)$ sufficiently close to the number $a$ .

The Left-Hand Limit of a Function

Suppose that $f$ is defined on an open interval (a,c).

L is the left-hand limit of $f(x)$ as $x$ approaches $a$ , and we have:

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

If $f(x)$ can be made as close to $L$ by chossing a point $x$ in $(c,a)$ sufficiently close to the number $a$ .

THEOREM:

If a function $f$ is defined on a deleted neighborhood of the point $a$ .

Then the limit of f(x) exists and is equal to the number $L$ if and only if the one-sided limits both exist and equal to $L$ .

Squeeze Law

If we assume that $f(x) \leqq g(x) \leqq h(x)$ in a deleted neighborhood of $a$ and also that:

$\lim_{x \to a} f(x)=L=\lim_{x \to a} h(x)$

We can write:

$\lim_{x \to a} g(x)=L$

This is very valuable in limits evaluation.

Continuity of functions:

If a function $f$ is defined in a neighborhood of $a$ , we can say that $f$ is continuous at $a$ if :

$\lim_{x \to a} f(x)$

exists and,

the value of the limit is $f(a)$

For a function to be continuous at a point $a$ :

– $f$ has to be defined at $a$ , meaning $f(a)$ exists;

-the limit of $f(x)$ as $x$ approaches $a$ must exist;

-the limit of $f(x)$ is equal to $f(a)$ .

Trigonometric Functions limits

We can easily show that:

$\lim_{\theta \to 0} \cos \theta=1$

$\lim_{\theta \to 0} \sin \theta=0$

The $sine$ and $cosine$ functions are continuous on the real line.

We can also prove:

$\lim_{\theta \to 0}\frac{ \sin \theta}{\theta}=1$

Post Views: 188

Limits and Continuity