$$ f(x,y) = x^\frac{1}{3}y^\frac{1}{3} $$

$$\frac{\partial f}{\partial x}(0,0) = \lim_{x \to 0} \frac{f(h,0)-f(0,0)}{h}= \lim_{x \to 0} \frac{0-0}{h} = 0$$

"and, similarly, $\frac{\partial f}{\partial y}(0,0) =0$ (these are not indeterminate forms!). It is necessary to use the original definition of partial derivatives, because the functions $x^\frac{1}{3}$ and $y^\frac{1}{3}$ are not themselves differentiable at 0."

This is a portion of textbook explaning why a simple definition of a partial derivative does not work but a linear approximation definition of a partial derivative must be used.

However, I'm confused at this part where they seem to be trying to use a counterexample to prove why a simple definition of partial derivatives does not work. Isn't this limit an indeterminate form? Yet, as you can see, the textbook claims this limit is not an indeterminate form to make their case.

I would greatly appreciate your help in making sense of this textbook.

Reference textbook: Vector Calculus by Marsden and Tromba 5th edition.

  • 4
    $\begingroup$ Shouldn't it be $\lim_{h \to 0}$ ? $\endgroup$
    – Digitallis
    Commented Jul 6, 2020 at 10:26

2 Answers 2


You have to first compute $\frac {f(h,0)-f(0,0)} h$ and then take the limit. Since $\frac {f(h,0)-f(0,0)} h=0$ for every $h \neq 0$ the limit is $0$. It is not an indeterminate form.


It is defined because the $0$ in the numerator of your limit is "real" $0$. Look, in the expression: $$\lim_{x\rightarrow 0}\frac{0}{x}$$ The answer is in fact $0$, because the denominator will be a very little value close to $0$, but not $0$, while the numerator is just $0$ ($0$ divided by any real number different to $0$ is equal to $0$, and the $x$ in the denominator isn't equal to $0$ despite being as close to it as you want). So in fact: $$\lim_{x\rightarrow 0}\frac{0}{x}=0.$$

So your limit isn't an indeterminate form.


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