Calculus/Formal Definition of the Limit

Formal definition of a limit

In preliminary calculus, the definition of a limit is probably the most difficult concept to grasp (If nothing else, it took some of the most brilliant mathematicians 150 years to arrive at it); it is also the most important and most useful.

The intuitive definition of a limit is adequate for manipulation most of the time, but is inadequate to understand the concept, or to prove anything with it. The issue here lies with our meaning of "arbitrarily close". We discussed earlier that the meaning of this term is that the closer x gets to the specified value, the closer the function must get to the limit, so that however close we want the function to the limit, we can find a corresponding x close to our value. We can express this concept as follows:

Definition: (Formal definition of a limit) Let f(x) be a function defined on an open interval that contains x=c, except possibly at x=c. Let L be an existing number. Then we say that, $\underset{x\to c}{\lim }f(x)=L$ if, for every $\epsilon >0$, there exists a $\delta >0$ such that for all x$\in$Df when $0<|x-c|<\delta ,$ we have $|f(x)-L|<\epsilon .$

To further explain, earlier we said that "however close we want the function to the limit, we can find a corresponding x close to our value." Using our new notation of epsilon ($ϵ$) and delta ($\delta$), we mean that if we want to find f(x) within $ϵ$ of L, the limit, then we know that there is a x within $\delta$ of c that puts it there.

Again, since this is tricky; let's resume our example from before: $f(x)=x^2$, at $x=2$. To start, let's say we want f(x) to be within .01 of the limit. We know here that the limit should be 4, so we say; for $ϵ=.01$, there is some delta so that as long as $0<|x-c|<\delta$, then $|f(x)-L|<\epsilon .$

To show this, we can pick ANY delta that is bigger than 0. To be sure, you might pick $.00000000000001$, because you are absolutely sure that if x is within .00000000000001 of 2, then f(x) will be within .01 of 4. Of course, we can't just pick a specific value for epsilon, like .01, because we said in our definition "for every $ϵ>0$." This means that we need to be able to show an infinite number of deltas, one for each epsilon. We can't list an infinite number of deltas!

Of course, we know of a very good way to do this; we simply create a function, so that for every epsilon, it can give us a delta. In this case, it's a rather easy function; all we need is $\delta (\epsilon )<\sqrt{(}\epsilon )$.

So how do you show that f(x) tends to L as x tends to c? Well imagine somebody gave you a small number $ϵ$ (e.g. say $ϵ=0.03$). Then you have to find a $\delta >0$ and show that whenever $0<|x-c|<\delta$ we have $|f(x)-L|<0.03$. Now if that person gave you a smaller $ϵ$ (say $ϵ=0.002$) then you would have to find another $\delta$, but this time with 0.03 replaced by 0.002. If you can do this for any choice of $ϵ$ then you have shown that f(x) tends to L as x tends to c.

Definition:(limit of a function at infinity) We call L the limit of f(x) as x approaches $\mathrm{\infty }$ if for every number $ϵ>0$ there exists a $\delta$; such that whenever $x>\delta$ we have

$|f(x)-L|<\epsilon$

When this holds we write

$\underset{x\to \mathrm{\infty }}{\lim }f(x)=L$

or

$f(x)\to Lasx\to \mathrm{\infty }.$

Similarly, we call L the limit of f(x) as x approaches $-\mathrm{\infty }$ if for every number $ϵ>0$, there exists a number $\delta$ such that whenever $x<\delta$ we have

$|f(x)-L|<\epsilon$

When this holds we write

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

or

$f(x)\to Lasx\to -\mathrm{\infty }.$

Notice the difference in these two definitions. For the limit of f(x) as x approaches $\mathrm{\infty }$ we are interested in those x such that $x>\delta$. For the limit of f(x) as x approaches $-\mathrm{\infty }$ we are interested in those x such that $x<\delta$.

Examples

Here are some examples on finding limits using the definition.

1) What is δ when ε = 0.01 for

$\underset{x\to 8}{\lim }\frac{x}{4}=2$?

we start with the conclusion and substitute the given values for f(x) and ε

$|\frac{x}{4}-2|<0.01$

and simplify

$7.96<x<8.04$

using the first part of the definition of a limit

$-0.04<x-8<0.04$

we normally choose the smaller of |-0.04| and 0.04 for δ but any smaller number will work so δ=0.04

2) What is the limit of f(x) = x + 7 as x approaches 4?

There are two steps to answering such a question; first we must determine the answer -- this is where intuition and guessing is useful, as well as the informal definition of a limit. Then, we must prove that the answer is right. For this problem, the answer happens to be 11. Now, we must prove it using the definition of a limit:

Informal: 11 is the limit because when x is roughly equal to 4, f(x) = x + 7 approximately equals 4 + 7, which equals 11.

Formal: We need to prove that no matter what value of ε is given to us, we can find a value of δ such that

$|f(x)-11|<\epsilon$

whenever

$|x-4|<\delta .$

For this particular problem, letting δ equal ε works (see choosing delta for help in determining the value of delta to use). Now, we have to prove

$|f(x)-11|<\epsilon$

given that

$|x-4|<\delta =\epsilon .$

Since |x - 4| < ε, we know
|f(x) - 11|  = |x + 7 - 11| = |x - 4| < ε, which is what we wished to prove.

3) What is the limit of f(x) = x² as x approaches 4?

Formal: Again, we pull two things out of thin air; the limit is 16 (use the informal definition to find the limit of f(x)), and δ equals √(ε+16) - 4. Note that δ is always positive for positive ε. Now, we have to prove
$|x^2-16|<\epsilon$
given that
$|x-4|<\delta =\sqrt{\epsilon +16}-4$.

We know that |x + 4| = |(x - 4) + 8| ≤ |x - 4| + 8  < δ + 8 (because of the triangle inequality), thus

$\begin{array}{ccc}|x^2-16|& =& |x-4|\cdot |x+4|\\ \\ & <& (\delta )\cdot (\delta +8)\\ \\ & <& (\sqrt{16+\epsilon }-4)\cdot (\sqrt{16+\epsilon }+4)\\ \\ & <& (\sqrt{16+\epsilon }{)}^2-4^2\\ \\ & <& \epsilon \end{array}.$

4) Show that the limit of sin(1/x) as x approaches 0 does not exist.

Suppose the limit exists and is l. We will proceed by contradiction. Assume that $l\ne 1$, the case for $l=1$ is similar. Choose $ϵ=|l-1|$, then for every $\delta >0$, there exists a large enough n such that $0<x_0=\frac{1}{\pi /2+2\pi n}<\delta$, but $|\sin (1/x_0)-l|=|1-l|<ϵ$ a contradiction.

The function sin(1/x) is known as the topologist's comb.

5) What is the limit of $x\sin (1/x)$ as x approaches 0?

It is 0. For every $ϵ>0$, choose $\delta =ϵ$ so that for all x, if $0<|x|<\delta$, then $|x\sin (1/x)-0|\le |x|<\epsilon$ as required. Can you prove that ln lim(x->a)f(x)=lim(x->a)lnf(x)?