Définition de limite
Cours : Définition de limite. Recherche parmi 300 000+ dissertationsPar Decimus Lucum • 24 Novembre 2020 • Cours • 1 148 Mots (5 Pages) • 662 Vues
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 =⇒
...