Définition de limite

The limit is formally defined as follows:

limx→a

f(x) = L

if for every number  > 0 there is a corresponding number δ > 0 such that

0 < |x − a| < δ =⇒ |f(x) − L| < 

Intuitively, this means that for any , you can find a δ such that |f(x)−L| < .

To do the formal  − δ proof, we will first take  as given, and substitute

into the |f(x) − L| <  part of the definition. Then we will try to manipulate

this expression into the form |x − a| < something. We will then let δ be this

“something” and then using that δ, prove that the  − δ condition holds. Some

examples should make this clear.

1. Prove:

limx→4

x = 4

We must first determine what a and L are. In this case, a = 4 (the value

the variable is approaching), and L = 4 (the final value of the limit). The

function is f(x) = x, since that is what we are taking the limit of.

Following the procedure outlined above, we will first take epsilon, as given,

and substitute into |f(x) − L| <  part of the expression:

|f(x) − L| <  =⇒ |x − 4| < 

In this case we are lucky, because the expression has naturally simplified

down to the form |x − a| < δ! Therefore, since we know from the above

that |x − 4| < , we can let δ = , and we know that |x − 4| < δ. This last

point is very important.

We can now finish the proof:

Given , let δ = . Then:

|x − a| < δ =⇒ |x − 4| < 

=⇒ |f(x) − L| < 

This completes the proof.

1

2. Prove:

limx→1

3x − 1 = 2

In this problem, a = 1, L = 2, f(x) = 3x − 1. We will proceed as we did

before, by substituting into |f(x) − L| < .

|f(x) − L| <  =⇒ |(3x − 1) − 2| < 

=⇒ |3x − 3| < 

=⇒ 3|x − 1| < 

=⇒ |x − 1| < /3

Once again, we have simplified the expression down to the form |x−a| < δ.

In this case, we can let δ = /3, and we can write up the proof: Given ,

let δ = /3. Then:

|x − a| < δ =⇒ |x − 1| < /3

=⇒ 3|x − 1| < 

=⇒ |3x − 3| < 

=⇒ |(3x − 1) − 2| < 

=⇒ |f(x) − L| < 

This completes the proof.

3. Prove:

limx→∞

√

x = ∞

In this problem, we have a = ∞ and L = ∞. If we try to apply the proof

directly, we will end up |f(x) − ∞| < , which produces a meaningless

result, since, anything minus ∞ is ∞. Therefore, we need to modify or

definition of limit slightly for infinity problems.

Let us first consider what it means for the limit to be equal to infinity.

When the limit equals infinity, then that means that f(x) increases without bound. In other words, for any positive integer M, there is a δ such

that within 0 < |x−a| < δ, f(x) > M (mathematically, this is saying that

as you approach the point a, f(x) becomes larger than any finite value.

Now let us consider the other side: x → a. If we directly substitute a = ∞,

following the same arguments as the last paragraph, we end up with the

same meaningless values. Therefore, we will also use a similar argument

to replace 0 < |x − a| < δ with x > N, where N is any positive integer.

Therefore, our modified definition of  − δ for this problem will be:

limx→∞

f(x) = ∞

if for every positive integer M there is a corresponding positive integer N

such that

2

x > N =⇒ f(x) > M

* You should work out for yourself, what the precise definitions of limit

would be if only one of x or f(x) is infinity, or if one of these is negative

infinity, etc.

Now we can proceed as before:

f(x) > M =⇒

...