I am really confused and was hoping someone could help clear up my understanding.

Suppose I have two continuous differentiable surfaces $f,g:\mathbb{R}^N\rightarrow\mathbb{R}$ and there exists a point $x_0$ such that $f(x_0)=g(x_0)=0$. Further $g(x)$ is non-zero everywhere else in the domain. I am interested in the surface $\frac{f}{g}$, which must be continuous everywhere where $g$ is non-zero.

Suppose I find a curve $C_1:[0,1]\rightarrow\mathbb{R}^N$ such that $x_0=C_1(0)$ and $x_1=C_1(1)$ for some $x_1 \in \mathbb{R}^N$ and I am able to show that: $$\lim_{y \rightarrow 0}\frac{f(C_1(y))}{g(C_1(y))}=k$$

The question which is confusing me is this:

Does the existence of the above limit imply that if I were to find any curve function, say $C_2:[0,1]\rightarrow\mathbb{R}^N$ such that $x_0=C_2(0)$ and $x_2=C_2(1)$ for some $x_2 \in \mathbb{R}^N$, it would be true that:

$$\lim_{y \rightarrow 0}\frac{f(C_2(y))}{g(C_2(y))}=k$$

i.e. is the limit now the same along all curves? Or in other words, if I defined $h(x)=\frac{f(x)}{g(x)}$ for all $x\neq x_0$ and $h(x_0)=k$ would $h$ be continuous?

My intuition for why the answer to the above should be "yes, $h$ would be continuous" is this:

  • $x_0$ should be a "removable" discontinuous point for $\frac{f}{g}$ and therefore it seems it should have a well-defined limit at $x_0$ from all sides and that limit should be equal.

My intuition for why the answer to the above should be "no, $h$ is not continuous" is this:

  • Suppose additionally that $C_1$ and $C_2$ were differentiable everywhere and that $f$ and $g$ had a non-zero directional derivative along $C_1$ and $C_2$. By L'hopitals rule, we would have: $$k=\lim_{y\rightarrow 0}\frac{(f \circ C_1) (y)}{(g \circ C_1) (y)}=\lim_{y\rightarrow 0}\frac{(f \circ C_1)' (y)}{(g \circ C_1)' (y)}=\lim_{x\rightarrow x_0}\frac{\nabla_{C_1} f(x)}{\nabla_{C_1} g(x)}$$ and: $$k=\lim_{y\rightarrow 0}\frac{(f \circ C_2) (y)}{(g \circ C_2) (y)}=\lim_{y\rightarrow 0}\frac{(f \circ C_2)' (y)}{(g \circ C_2)' (y)}=\lim_{x\rightarrow x_0}\frac{\nabla_{C_2} f(x)}{\nabla_{C_2} g(x)}$$ Thus: $$\lim_{x\rightarrow x_0}\frac{\nabla_{C_1} f(x)}{\nabla_{C_1} g(x)}=\lim_{x\rightarrow x_0}\frac{\nabla_{C_2} f(x)}{\nabla_{C_2} g(x)}$$ This seems like quite a strong statement and its hard for me to believe it is true for every curve $C_2$ such that $g$ has non-zero directional derivative along $C_2$...

Apologies for the very long post but I am really very confused and any help clearing up my understanding here would be very appreciated!


If I understand your question correctly, the claim is clearly false. Just take $f(x,y)=xy$ and $g(x,y)=x^2+y^2$ (with $N=2$, obviously), so that $$ h(x,y) = \frac{xy}{x^2+y^2} . $$ For $C_1(t)=(t,0)$ the limit of $h(t,0)$ as $t \to 0$ is zero, and for $C_2(t)=(t,t)$ the limit of $h(t,t)$ is $1/2$. The confusion seems to lie in your idea about a removable singularity, which isn't correct.

  • $\begingroup$ Ah! Right I see yes thank you $\endgroup$
    – JDoe2
    Mar 10 at 14:57

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