Using L'Hospital's rule to evaluate a fraction containing two variables

[seeker101]

This might be a silly question:

Is there a meaningful way to use L'Hospital's rule to determine
$$\lim_{\vec{\mathbf{x}} \rightarrow \vec{\mathbf{0}}}\frac{x^3}{x^2+y^2}?$$

where
$$\vec{\mathbf{x}}=\left(\begin{array}{c}x\\y\end{array}\right)$$

 

[Hootenanny]

HINT: Are you sure that the limit exists?

 

[seeker101]

I think the limit does exist (I am unable to see any reasons for its non-existence; isn't zero the limit?).

But my question is really about using L'Hospital's rule to evaluate
$$\lim_{\vec{\mathbf{x}} \rightarrow \vec{\mathbf{0}}}\frac{f(x,y)}{g(x,y)}$$

assuming the numerator and denominator vanish as
$$\left(\begin{array}{c}x\\y\end{ar ray}\right)\rightarrow\left(\begin{array}{c}0\\0\end{ar ray}\right)$$

 

[Hootenanny]

I think the limit does exist (I am unable to see any reasons for its non-existence; isn't zero the limit?).

But my question is really about using L'Hospital's rule to evaluate
$$\lim_{\vec{\mathbf{x}} \rightarrow \vec{\mathbf{0}}}\frac{f(x,y)}{g(x,y)}$$

assuming the numerator and denominator vanish as
$$\left(\begin{array}{c}x\\y\end{ar ray}\right)\rightarrow\left(\begin{array}{c}0\\0\end{ar ray}\right)$$

The limit does indeed exists. However, I was hinting that if the limit does exist, then it must have the same value irrespective of the path you take

 

[seeker101]

Thank you for responding. But finding the limit is not my objective. I was wondering if L'Hospital's rule is even applicable in a situation as above (two variable case).

 

[Hootenanny]

Thank you for responding. But finding the limit is not my objective. I was wondering if L'Hospital's rule is even applicable in a situation as above (two variable case).

Yes, one may use L'Hopital's rule for multi-variable functions. However, one must use the directional derivative as opposed to the standard derivative. The directional derivative is taken in the direction of the limit.

However, in the example which you have quoted, there is no need to make use of L'Hopital's rule.