Lim a_n = L => lim sqrt(a_n) = sqrt(L)

  • Thread starter Damidami
  • Start date
In summary: For example, if you could make it bigger by multiplying it by a bigger number. Or maybe taking the square root of the larger number. These are all things that can be done to the term to make it bigger.
  • #1
94
0
Let [tex]\{a_n\}[/tex] be a sequence of positive numbers. If [tex]\lim a_n = A[/tex], prove that [tex]\lim \sqrt{a_n} = \sqrt{A}[/tex].

I thoght that [tex]f(x) = \sqrt{x}[/tex] is a continuous function, and composing a sequence that has a limit with a continuous function must have a limit that is just evaluating the previous limit in the function.

But I think I'm not supposed to use that fact, as the concept of limit of sequence comes previously to the definition of continuous function in the course.

It seems trivial, but still can't find how to start. Any help?
 
Physics news on Phys.org
  • #2
What is your definition of a limit?
 
  • #3
kru_ said:
What is your definition of a limit?

Hi kru_,
Isn't this standard for sequences?

A sequence [tex]a_n[/tex] of real numbers has limit [tex]L \in \mathbb{R}[/tex] iff for every [tex]\epsilon > 0[/tex] there exists [tex]n_0 \in \mathbb{N}[/tex] such that for all [tex]n \geq n_0[/tex] one has [tex]|a_n - L| < \epsilon[/tex]
 
  • #4
So you know [itex]|a_n - A| < \epsilon[/itex].

I think the key hint here is that [itex]|\sqrt{a_n} - \sqrt{A}| = \frac{|a_n - A|}{|\sqrt{a_n} + \sqrt{A}|}[/itex]
 
  • #5
Thank you, but I still can't see how to end it. Most that can see is that

[itex] | \sqrt{a_n} - \sqrt{A}| < \frac{\epsilon}{|\sqrt{a_n} + \sqrt{A}|} [/itex]

And I'm not sure if it's still valid if [itex] \lim_{n \to \infty} a_n = 0[/itex]
 
  • #6
Damidami said:
Thank you, but I still can't see how to end it. Most that can see is that

[itex] | \sqrt{a_n} - \sqrt{A}| < \frac{\epsilon}{|\sqrt{a_n} + \sqrt{A}|} [/itex]

And I'm not sure if it's still valid if [itex] \lim_{n \to \infty} a_n = 0[/itex]

You are right that you will need to handle a separate case if the limit is 0. This should be a pretty straightforward case, however. What does the definition of the limit look like if A = 0?


Focusing on the general case, where A is not 0:

You know that [itex]|a_n - A| \leq \epsilon[/itex]. The crucial component of this knowledge is that this remains true for all positive epsilon. It remains true for [itex]2\epsilon, 3\epsilon, \frac{3\epsilon}{7}, 4\pi\epsilon, \epsilon^2,..., A\epsilon^2, \sqrt{A}\epsilon[/itex], etc.

Does this make sense?


Another thing to consider is what you can do to the term [itex]\frac{|a_n - A|}{|\sqrt{a_n} + \sqrt{A}|}[/itex] to make it bigger.
 

What is the meaning of "lim a_n = L"?

This notation represents the limit of a sequence a_n as n approaches infinity, where the limit is equal to the value L.

How is this notation related to the statement "lim sqrt(a_n) = sqrt(L)"?

This notation states that the limit of the square root of the sequence a_n is equal to the square root of the limit L.

What is the significance of the square root in this statement?

The square root is significant because it represents a mathematical operation that is applied to both the sequence a_n and the limit L.

What is the intuition behind this statement?

This statement suggests that taking the square root of a sequence and its limit will result in the same value as taking the limit of the original sequence. In other words, the square root operation does not affect the limit of a sequence.

Are there any restrictions on the values of a_n and L for this statement to hold true?

Yes, for this statement to hold true, the sequence a_n must approach a finite limit L as n approaches infinity. Additionally, both the sequence and the limit must have non-negative values to ensure the existence of the square root.

Suggested for: Lim a_n = L => lim sqrt(a_n) = sqrt(L)

Replies
5
Views
693
Replies
8
Views
910
Replies
1
Views
674
Replies
9
Views
911
Replies
24
Views
987
Back
Top