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

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Homework Help Overview

The discussion centers around the limit of a sequence of positive numbers, specifically exploring the relationship between the limit of the sequence \{a_n\} and the limit of its square root, \sqrt{a_n}. The original poster attempts to prove that if \lim a_n = A, then \lim \sqrt{a_n} = \sqrt{A}, while expressing uncertainty about the use of continuity in their reasoning.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the definition of a limit and its implications for sequences. There is a focus on the relationship between the limits of a sequence and its transformation through the square root function. Questions arise about handling cases where the limit approaches zero and the validity of certain inequalities.

Discussion Status

The discussion is ongoing, with participants providing hints and exploring different aspects of the problem. Some guidance has been offered regarding the manipulation of inequalities and the need to consider special cases, particularly when the limit is zero. However, there is no explicit consensus on the final approach or solution.

Contextual Notes

Participants note the importance of the definition of limits in their reasoning and the potential complications that arise when the limit approaches zero. There is an acknowledgment of the need to handle different cases separately, which adds complexity to the discussion.

Damidami
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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?
 
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What is your definition of a limit?
 
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]
 
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]
 
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]
 
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.
 

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