Find $f$ and $g$ that are both $C^1$ with $f(y)=g(y)=0$ and $f'(y)= a g'(y) \neq 0$, but $\lim_{x\to y} \frac{f(x)}{g(x)}$ DNE.

Question

Find $f$ and $g$ that are both $C^1$ with $f(y)=g(y)=0$ and $f'(y)= a g'(y) \neq 0$, where $a\in\mathbb{R}$ but $\lim_{x\to y} \frac{f(x)}{g(x)}$ DNE.

I know that for this to be true we must have that $f,g : \mathbb{R}^n \to \mathbb{R}$, where $n \neq 1$.

Answer 1

Let's work in $\mathbb R^2.$ Take $y=(0,0).$ Set

$$f(x_1,x_2) = x_1 +x_1^2,\,\, g(x_1,x_2) = x_1 +x_2^2.$$

We have the hypotheses met with $\nabla f(0,0) = (1,0)=\nabla g(0,0),$ and $a=1.$

Note $f(0,t)/g(0,t) = 0/t^2 \to 0$ as $t\to 0,$ while $f(t,0)/g(t,0) \to 1$ as $t\to 0.$ Thus $\lim f(x)/g(x)$ fails to exist as $x\to (0,0).$

Answer 2

Two possibly interesting new bits: (i) for real-valued functions on $\Bbb R^n$ we don't need to construct a counterexample, in fact there are no non-counterexamples. (ii) There is a sort of L'Hopital for functions from $\Bbb R^n$ to itself.

First,

If $n>1$, $f,g:\Bbb R^n\to\Bbb R$, $g(0)=0$, $g'(0)$ exists and $g'(0)\ne0$ then $\lim_{x\to0}f(x)/g(x)$ does not exist.

This is clear from

If $g$ is as above and $\delta>0$ then there exists $x$ with $0<|x|<\delta$ and $g(x)=0$.

Proof: Let $A=\{0<|x|<\delta\}$. Since $g'(0)\ne0$ there exist $x,y\in A$ with $g(x)<0$ and $g(y)>0$. The fact that $n>1$ shows that $A$ is connected, and so $g$ vanishes at some point of $A$.

Second: Of course the very notion of L'Hopital for $f,g:\Bbb R^n\to\Bbb R^n$ seems absurd, since there's no such thing as $f/g$. But we can rephrase things to avoid division: For real-valued functions saying $f/g\to\alpha$ is the same as saying $$f-\alpha g=o(g).$$

It turns out that the condition $g'(0)\ne0$ is not right; what's relevant is whether $g'(0)$ is invertible:

Suppose $f,g:\Bbb R^n\to \Bbb R^n$, $f(0)=g(0)=0$, $f$ and $g$ are differentiable at the origin, $g'(0)$ is invertible and $f'(0)=\alpha g'(0)$. Then $f(x)-\alpha g(x)=o(g(x))$ as $x\to0$.

Proof: Say $A=g'(0)$, so we have $$g(x)=Ax+o(x)$$and$$f(x)=\alpha Ax+o(x).$$Hence $$f(x)-\alpha g(x)=o(x)\quad(x\to0).$$

But since $A$ is invertible there exist $c>0$ and $\delta>0$ with $$|g(x)|\ge c|x|\quad(|x|<\delta).$$So, perhaps a bit iinformally, $$o(x)=o(g(x)).$$

Original, where I assumed we were talking about functions from $\Bbb R$ to $\Bbb R$:

No, L'Hopital's rule does not fail, not ever. It's a theorem. The conclusion may fail, if the hypotheses are not satisfied.

Anyway, it's easy to see that the example you ask for cannot exist. In fact

If $f(y)=g(y)=0$, $f'(y)$ and $g'(y)$ exist, and $g'(y)\ne0$ then $f(x)/g(x)\to f'(y)/g'(y)$ as $x\to y$.

This is trivial from the definition of the derivative. Note first that since $g'(y)\ne0$ there exists $\delta>0$ such that $g(x)\ne0$ for $0<|x-y|<\delta$. For such $x$ we have $$\frac{f(x)}{g(x)}=\frac{(f(x)-f(y))/(x-y)}{(g(x)-g(y))/(x-y)}\to \frac{f'(y)}{g'(y)}\quad(x\to y).$$