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Evaluate the limit $$\displaystyle \lim_{(x,y) \to (0,0)}\frac{\cos(x) - 1 - \frac{x^2}{2}}{x^4 + y^4}$$

I know to evaluate limits of multiple variables, I have to check if the limits along different paths are the same. If they are different, the limit does not exist because that would mean the path do not 'tend' to the same thing.

In my limit problem, I noticed that with the paths $x = y$, $y=x$, $(x,0)$, and $(0,y)$ I will always end up with an indeterminate form. What does this mean? Usually I would continue with L'Hopital's rule, but I was told that I cannot do that with multivariable limits. However, could I not use L'Hopital's rule when I hold $x$ or $y$ constant? I would only be working with a single variable at that point.

Robert Z
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Ozera
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1 Answers1

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Let $R(x,y)$ denote the value of the function at $(x,y)\ne(0,0)$.

When $y\to0$, $\cos(y^2)-1-\frac12y^4\sim-y^4$ and $(y^2)^4+y^4\sim y^4$ hence $R(y^2,y)\to-1$ and $R(0,y)=0\to0$ when $y\to0$. Thus the limit of $R(x,y)$ when $(x,y)\to(0,0)$ does not exist.

That $R$ has a limit at $(0,0)$, say the limit $\ell$, would mean that for every positive $\varepsilon$, there exists $\delta$ such that $$ 0\lt x^2+y^2\lt\delta\implies|R(x,y)-\ell|\lt\varepsilon. $$ When $y\to0$, $(y^2,y)\to(0,0)$ and $(0,y)\to(0,0)$, hence, one would have $R(y^2,y)\to\ell$ and $R(0,y)\to\ell$ when $y\to0$. Since $-1\ne0$, $R$ has no limit at $(0,0)$.

(Unfortunately, I have no idea how one would "continue with L'Hopital's rule", not even why one would summon this rule in such a context.)

Did
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  • the numerator is $\cos(x)-1-\dfrac12x^2$ – obataku Sep 06 '13 at 20:33
  • @oldrinb It seems so. Thanks (see modified version). – Did Sep 06 '13 at 20:44
  • @Did I don't think I have ever seen this method before. Is there no other way to conclude that this limit does not exist? Finding that the paths I found result in indeterminate forms doesn't mean anything right? That isn't enough information to conclude that the limit does not exist right? – Ozera Sep 06 '13 at 20:59
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    Actually the limits along the paths $y=x$ and $y=0$ are $-\infty$ and the limit along the path $x=0$ is $0$ hence this works as well. – Did Sep 06 '13 at 21:04
  • Ah! Yes I computed the limits incorrectly for $y=x$. Thanks! – Ozera Sep 06 '13 at 21:10