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Give an example of continuously differentiable functions $f$,$g:\mathbb{R}^n\to\mathbb{R}$ for which $f(x_0)=g(x_0)=0$ and $\nabla f(x_0) = \lambda \nabla g(x_0) \neq 0$, for some point $x_0$ and number $\lambda$, but $$\lim\limits_{x\to x_0} \frac{f(x)}{g(x)}$$ does not exist.

Proof

Let the $i$th component of $g(x)$ be given by $g(x_i) = \frac{1}{x_{0_i}}(x_i^2-x_{0_i}^2)$.

Let the $i$th component of $f(x)$ be given by $f(x_i) = x_i-x_{0_i}$.

Then $f(x_0) = g(x_0) = 0$ and $\nabla g(x_0) = 2$, and $\nabla f(x_0) = 1$.

However, this limit is identical to $\lim\limits_{x\to 0} \frac{1}{x}$, which is known to not exist.

Something about this proof seems invalid to me.

MurderOfCrows
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    Only thing I see wrong is that it is the components of the gradients that are $2$ and $1$ respectively. That and the fact that you need to assume that all components of $x_0$ are nonzero for this example, but that's OK because it just needs to be a particular example of failure. – Ian Mar 20 '20 at 23:53
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    @Ian, What does the OP mean by "the $i$th component of $g(x)$"? There is an $i$th component of $x\in\mathbb{R^n}$, but $g(x)$ is a single real number. Same for $f(x)$. – Barry Cipra Mar 20 '20 at 23:57
  • What is your version of L'Hospital's rule in $\mathbb{R}^n$? I've seen generalizations to $\mathbb{R}^2$ only but even that is not a very popular theorem. – bjorn93 Mar 21 '20 at 00:46
  • @BarryCipra Maybe they mean to say $g(x)=\sum_{i=1}^n \frac{x_i^2-(x_0)i}{(x_0)_i}$ and $f(x)=\sum{i=1}^n x_i - (x_0)_i$? – Ian Mar 21 '20 at 01:21

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