Evaluating a limit with l'hopital's rule

In summary: You can't use it to prove that it does have a limit. So you still need to know how to do that.In summary, the conversation discusses evaluating a limit of a particular form in the context of a physics problem. The original approach of using l'Hopital's rule is not successful due to circular derivatives, and alternative methods such as using polynomials or dividing by x may be used to evaluate the limit. The conversation also touches on the limitations of l'Hopital's rule and the importance of first proving that a limit exists before applying it.
  • #1
lets_resonate
15
0
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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  • #2
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!
 
  • #3
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.
 
  • #4
HINT: As x becomes large, one can negate the constant :wink:
 
  • #5
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?
 
  • #6
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.
 
  • #7
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:

What is l'hopital's rule?

L'Hopital's rule is a mathematical tool used to evaluate limits of indeterminate forms, where both the numerator and denominator approach zero or infinity.

When should l'hopital's rule be used?

L'Hopital's rule should only be used when evaluating limits of indeterminate forms, such as 0/0 or ∞/∞. It cannot be used for limits with other types of indeterminate forms, such as 0*∞ or ∞-∞.

How is l'hopital's rule applied?

To apply l'Hopital's rule, take the derivative of the numerator and denominator separately, and then evaluate the limit again. If the new limit is still indeterminate, repeat the process until a non-indeterminate form is obtained.

What are the limitations of l'hopital's rule?

L'Hopital's rule can only be used for evaluating limits of indeterminate forms. It cannot be used for evaluating limits at infinity or for finding the value of a limit if it exists. It also assumes that the functions involved are differentiable.

Can l'hopital's rule be used for all types of functions?

No, l'Hopital's rule can only be used for functions that are differentiable. If a function is not differentiable, l'Hopital's rule cannot be applied to evaluate its limit.

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