0

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?

Martin Sleziak
  • 53,687
Sha Vuklia
  • 3,960
  • 4
  • 19
  • 37
  • 1
    No, you have not interpreted anything incorrectly. While it does not technically matter, the expression you have written will appear first in logical order when we proceed by definition. – Sarvesh Ravichandran Iyer Mar 06 '17 at 08:58
  • 2
    @Sha I think you are right. The author probably confused the definition with the expressions for $;f_x,,f_y;$ that he wrote just a few lines above... – DonAntonio Mar 06 '17 at 08:59
  • 1
    The author probably confused nothing with nothing, instead, they simply were aware that $h/|h|$ and $|h|/h$ are the same. – Did Mar 06 '17 at 09:12

1 Answers1

0

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 $0$.

Serge Ballesta
  • 1,004