Evaluating a limit with l'hopital's rule

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Homework Help Overview

The discussion revolves around evaluating the limit \(\lim_{x \to \infty} \left( \frac{x}{\sqrt{x^2+y^2}} \right)\), which is related to calculus and the application of l'Hopital's rule. Participants are exploring the implications of the limit's form and the behavior of the expression as \(x\) approaches infinity.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of l'Hopital's rule and its limitations in this context, noting circular derivatives. Questions are raised about the proper approach to evaluate the limit and the implications of treating \(y\) as a constant versus a variable.

Discussion Status

The discussion is ongoing, with participants offering hints and alternative perspectives on evaluating the limit. Some suggest that dividing the numerator and denominator by \(x\) may provide clarity, while others question the assumptions made about the variables involved.

Contextual Notes

There is a recognition that the limit's evaluation may differ based on the treatment of \(y\) as a constant or variable, and that the circular nature of the derivatives presents a challenge in finding a resolution.

lets_resonate
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It's been a while since I've evaluated limits, and I'm beginning to forget some of the techniques. A problem came up in physics which involved evaluating a limit of this particular form.

Homework Statement



[tex]\lim_{x \to \infty} \left( \frac{x}{\sqrt{x^2+y^2}} \right)[/tex]

Homework Equations



L'hopital's rule would be my first guess at the proper approach.

The Attempt at a Solution



I know the limit will evaluate to be a 1. This could also be inferred from the fact that x is in the same degree in both the numerator and the denominator. But I tried to do it a bit more rigorously with l'Hopital's rule. The problem is that the expression inside the limit becomes circular with successive derivatives. That is, if we let [itex]f(x) = x[/itex] and [itex]g(x)=\sqrt{x^2+y^2}[/itex], we will find that:

[tex]\frac{f'(x)}{g'(x)}=\frac{g(x)}{f(x)}[/tex]

This will bring us no closer to finding the limit. Take the derivative of the numerator and the denominator again to find:

[tex]\frac{f''(x)}{g''(x)}=\frac{f(x)}{g(x)}[/tex]

Hey, we're back! L'Hopital took us for a spin and brought us back to the starting point.

So my questions:

1. What is the proper approach to evaluate the limit?

2. For the sake of curiosity, are there any interesting observations to be made about the situation? Perhaps a name for the circular nature of the problem?

Thanks in advance for any help.
 
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I mean technically you're in a correct form of infinity over infinity, but thinking broadly: I learned l'hopistal's rule in single variable calculus. You're writing g(x,y) as g(x) by mistake (at least I think so). I don't know if you need to use partials in three space as such, but I always remembered that limits were expressed as, say, x,y -> 0. The answer to your limit is probably different if y->0 vs. y-> infinity. Sorry for the ad-hoc suggestions, I'm not really sure!
 
Oh, sorry, in the context of the problem, [itex]y[/itex] is a constant, so putting [itex]y^2[/itex] is a little misleading. A more accurate statement of the problem would then be:

[tex] \lim_{x \to \infty} \left( \frac{x}{\sqrt{x^2+c}} \right)[/tex]

This doesn't change a lot. We still get those circular derivatives.
 
HINT: As x becomes large, one can negate the constant :wink:
 
lets_resonate said:
[tex]\frac{f'(x)}{g'(x)}=\frac{g(x)}{f(x)}[/tex]

This will bring us no closer to finding the limit.
Sure it does -- if L is the value of the original limit, then what are the limits of the left and right hand sides of that equation?


1. What is the proper approach to evaluate the limit?
The method you wanted to use in the beginning is one "proper" approach. Didn't you learn how to do it rigorously with polynomials? Won't the exact same method work here?
 
l'Hopital is not guaranteed to resolve a limit. In some cases, depending on how you write the indeterminant form, it can actually make things worse. You found one where it doesn't help. There is another way, divide numerator and denominator by x and move the 1/x inside the radical.
 
Hurkyl said:
Sure it does -- if L is the value of the original limit, then what are the limits of the left and right hand sides of that equation?

Good point. You can salvage something from l'Hopital. IF you assume it has a limit.
 
Last edited:

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