# Confusion over continuity on surfaces and L'Hospital's rule

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.