How to use L'hospital rule to compute the limit of the given function

$$\lim_{(x,y)\to (0,0)} \frac{x^{2}+y^{2}}{x+y}?$$

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    $\begingroup$ L'Hôpital's rule is strictly a one-variable thing. Inspect the given expression carefully, and you will detect what is going on here. $\endgroup$ – Christian Blatter Jul 31 '12 at 8:00

There is no L'Hopital's Rule for multiple variable limits. For calculating limits in multiple variables, you need to consider every possible path of approach of limits.

What you can do here:

Put $x=r\cos\theta$ and $y=r\sin\theta$, (polar coordinate system) and $(x,y)\to (0,0)$ gives you the limits $r\to 0$ and no limits on $\theta$.

Now we need to substitute these in your problem which then becomes $$\lim_{r\to 0}\dfrac{r^2}{r(\cos\theta+\sin\theta)}=\lim_{r\to 0}\dfrac{r}{(\cos\theta+\sin\theta)}$$ Now for paths where $\cos\theta\to-\sin\theta$ or $\cos\theta=-\sin\theta$, the denominator $\to0$ or $=0$ while the numerator is just tending to $0$ but not exactly $0$.Since the second case exists $\implies$ the limit doesn't exist.

If extra condition that $(x+y)\neq 0$ is applied, even then the limit does not exist as for path $\theta=0$, limit is $0$ while for path $\sin\theta=-\cos\theta+r\cos^2\theta$, limit is something which can be non-zero.

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    $\begingroup$ There are a lots of paths where $(x+y)=0$ many times so the limit is not defined... $\endgroup$ – Xoff Jul 31 '12 at 8:22
  • $\begingroup$ In the OP it should be noticed under the limit that $x+y \neq 0$. Besides, $(x,y) \rightarrow (0,0)$ depends on the metric you are using. When not specified it is the euclidian one so the limit actually means $||(x,y)||_2\rightarrow (0,0)$ which is equivalent to $r \rightarrow 0$. Thus any $\theta$ can be chosen to compute the limit. Eventually choose any but $\frac{3\pi}{4}$ or $-\frac{\pi}{4}$ mod $\pi$, which are forbidden by $x+y \neq 0$. $\endgroup$ – vanna Jul 31 '12 at 8:32
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    $\begingroup$ It's still wrong. See my answer. $\endgroup$ – Christian Blatter Jul 31 '12 at 9:59
  • $\begingroup$ Adding the condition $\,x+y\neq 0\,$ doesn't solve anything as there're lots of different points on the plane with the same polar angle...I think Christian's answer (below) pretty much clears this out. $\endgroup$ – DonAntonio Jul 31 '12 at 11:11
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    $\begingroup$ Notice this answer is only useful for proving that the limit does not exist. It cannot be used to prove that it does. For example, consider $f(x,y)=\left\{\begin{array}{lr} \frac{y\sqrt{x^2+y^2}}{x} & : x \neq0\\ y & : x = 0 \end{array}\right.$. All paths where $\theta$ is constant give the limit as $0$ but it is actually undefined. (The path $y=\sqrt{x}$ gives the limit to be $1$.) $\endgroup$ – Jay Jan 22 '14 at 1:31

The answer by Christian Blatter is correct. This particular limit is undefined.

I have recently written a paper giving a l'Hospital's rule for many types of multivariable limits. You can find it at



The limit doesn't exist, because every neighborhood $U_\epsilon({\bf 0})$ contains points $(x,y)$ where the expression $${x^2+y^2\over x+y}$$ is undefined.

Now one could be tempted to exclude the points $(x,y)$ with $x+y=0$ from consideration, but this is of no help: Let $S$ be the plane minus these points and consider an arbitrary sequence $x_n\to 0$ $\,(n\to\infty)$, $\, x_n>0$. Choosing $y_n:=-x_n+x_n^2$ ensures $(x_n,y_n)\in S$ and $(x_n,y_n)\to0$ $\,(n\to\infty)$. For such a sequence $${x_n^2+y_n^2 \over x_n+y_n}={2x_n^2+2x_n^3+x_n^4\over x_n^2}\to 2\qquad(n\to\infty)\ ,$$ whereas for the sequence $(x_n,0)$ the corresponding limit is $=0$.


Since there are two variables, the equation is a surface (2 dimensional) therefore, the limit is also a 2 dimensional patch. You should apply hopital rule in both directions, but since the equation is symmetric, that shouldn't be a problem.

  • $\begingroup$ Can you please elaborate? $\endgroup$ – Kns Jul 31 '12 at 7:28
  • $\begingroup$ You can first apply L'hopital rule in x direction, and leave y as is, (as a variable) Next do the same in y direction. $\endgroup$ – wesley Jul 31 '12 at 7:37
  • $\begingroup$ I don't think this is correct as this would imply we can iterate limits, which in the general case is wrong. $\endgroup$ – DonAntonio Jul 31 '12 at 11:04

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