Partial derivative at $(0,0)$ for norm function $f(x,y)=\sqrt{x^2+y^2}$

128 Views Asked by At

See example (ii):

enter image description here

Shouldn’t it be that we have a limit of the quotient $\frac{\mid h\mid}{h}$ instead of $\frac{h}{\mid h\mid}$? Because we have:

$$ \lim_{h\to 0}\frac{f(h,0)-f(0,0)}{h}=\lim_{h\to0}\frac{\sqrt{h^2}}{h}=\frac{\mid h\mid}{h}. $$

Technically, it doesn't matter of course, but I was just wondering if I interpreted something wrong?

1

There are 1 best solutions below

0
On BEST ANSWER

First of all $\dfrac{h}{\left| h \right|}$ is the same as $\dfrac{\left| h \right|}{h}$. More precisely: $$\forall h > 0, \dfrac{h}{\left| h \right|} = \dfrac{\left| h \right|}{h} = 1 \\ \forall h < 0, \dfrac{h}{\left| h \right|} = \dfrac{\left| h \right|}{h} = -1$$

The author simply says that as the right limit is distinct from the left limit, we cannot have a limit at $0$.