Proving the limit of an integral function

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Homework Help Overview

The problem involves evaluating the limit of the function R(x)/x^2 as x approaches infinity, where R(x) is defined as the integral from 0 to x of the function sqrt(1+t^2). Participants are tasked with finding this limit analytically rather than through numerical evaluation.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Some participants suggest using L'Hospital's rule to differentiate the integral and evaluate the limit.
  • Others propose starting with the indefinite integral before applying the limit.
  • There is discussion about the indeterminate form encountered when applying L'Hospital's rule and whether factoring out constants affects the limit.
  • Participants question the validity of certain steps and the implications of infinity in the context of limits.

Discussion Status

The discussion is ongoing, with various methods being explored, including L'Hospital's rule and alternative approaches to evaluate the limit. Participants are questioning assumptions and clarifying steps, indicating a productive exchange of ideas without reaching a consensus.

Contextual Notes

There is a focus on the behavior of the function as x approaches infinity, with participants expressing uncertainty about the best method to apply and the implications of their calculations. The conversation reflects a mix of approaches and interpretations regarding the limit and the use of L'Hospital's rule.

armolinasf
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Homework Statement


Show that the Lim x=> inf. R(x)/x^2 exists and find its value.


Homework Equations



R(x)= [tex]\int^{x}_{0}[/tex] [tex]\sqrt{1+t^{2}}[/tex]


The Attempt at a Solution



I know that the limit is 1/2 however i got that by just evaluating at large values of x. How would I go about solving this analytically? Thanks for the help
 
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Use L'Hospital's rule remembering how to differentiate an integral as a function of its upper limit.
 
how about starting by attempting the indifinite integral, then applying the limit
 
LCKurtz said:
Use L'Hospital's rule remembering how to differentiate an integral as a function of its upper limit.

much better idea, once you know the intergal is unbounded in the limit
 
Let me know if this works: applying L'H I get x/(1+x^2) * 1/2 if I set x equal to infinity wouldn't that give me inf/inf=1*1/2 which would give me the correct limit?
 
armolinasf said:
Let me know if this works: applying L'H I get x/(1+x^2) * 1/2 if I set x equal to infinity wouldn't that give me inf/inf=1*1/2 which would give me the correct limit?

That's not very clear. What do you get from l'Hopital? Shouldn't there be sqrt in there somewhere?
 
you're right, my mistake.

taking derivatives I get sqrt(1+x^2)/2x. So, can't I factor out 1/2 leaving me with sqrt(1+x^2)/x * 1/2. If I take the limit it should be infinity/infinity times 1/2 which is the same 1*1/2?
 
armolinasf said:
you're right, my mistake.

taking derivatives I get sqrt(1+x^2)/2x. So, can't I factor out 1/2 leaving me with sqrt(1+x^2)/x * 1/2. If I take the limit it should be infinity/infinity times 1/2 which is the same 1*1/2?

Don't EVER say infinity/infinity=1. That's the whole point of l'Hopital. infinity/infinity can be anything. Can't you prove the limit of sqrt(1+x^2)/x is 1 without saying that?
 
sqrt(1+x^2)/x is indeterminate when x=inf. So if I were apply l'Hopital again I get into this repeating pattern where I wind up at the same thing. For example, applying L'H to sqrt(1+x^2)/x gives x sqrt(1+x^2)/x and I'm back where I started.

By the way, the actual equation is sqrt(1+x^2)/2x I'm wondering if it's possible to factor out a 1/2 and then apply L'H after that? Or would that change the expression?
 
  • #10
armolinasf said:
sqrt(1+x^2)/x is indeterminate when x=inf. So if I were apply l'Hopital again I get into this repeating pattern where I wind up at the same thing. For example, applying L'H to sqrt(1+x^2)/x gives x sqrt(1+x^2)/x and I'm back where I started.

By the way, the actual equation is sqrt(1+x^2)/2x I'm wondering if it's possible to factor out a 1/2 and then apply L'H after that? Or would that change the expression?

Don't apply l'Hopital. (1/x)*sqrt(1+x^2)=sqrt((1/x^2)*(1+x^2)). l'Hopital isn't always the best way to do it. But sure you can factor out the 1/2.
 
  • #11
If I were to go at it without l'Hopital I would try to factor the numerator and the denominator and cancel out one of the x's but I'm not really seeing a way to factor sqrt(1+x^2)/x^2
 

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