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

 

[LCKurtz]

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

 

[lanedance]

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

 

[lanedance]

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?

 

[Dick]

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?

 

[Dick]

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?

 

[Dick]

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