# Proving the limit of an integral function

armolinasf

## Homework Statement

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

## Homework Equations

R(x)= $$\int^{x}_{0}$$ $$\sqrt{1+t^{2}}$$

## 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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## Answers and Replies

Homework Helper
Gold Member
Use L'Hospital's rule remembering how to differentiate an integral as a function of its upper limit.

Homework Helper
how about starting by attempting the indifinite integral, then applying the limit

Homework Helper
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

armolinasf
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?

Homework Helper
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?

armolinasf
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?

Homework Helper
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?

armolinasf
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?

Homework Helper
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.

armolinasf
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