MHB The Derivative in Several Variables .... Hubbard and Hubbard, Section 1.7 ....

[Math Amateur]

I am reading the book: "Vector Calculus, Linear Algebra and Differential Forms" (Fourth Edition) by John H Hubbard and Barbara Burke Hubbard.

I am currently focused on Section 1.7: Derivatives in Several Variables as Linear Transformations ...

I need some help in order to understand some remarks by Hubbard and Hubbard on page 124 under the heading "The derivative in several variables ... ...

The relevant text reads as follows:
View attachment 8720
Referring to equation 1.7.10 H&H say the following:

" ... ... But this wouldn't work even in dimension $$1$$, because the limit changes sign as $$h$$ approaches $$0$$ from the left and from the right. ... ... "Could someone please explain exactly how/why the limit changes sign as $$h$$ approaches $$0$$ from the left and from the right. ... ... ?

I am puzzled because $$\mid h \mid$$ doesn't change sign and $$f( a + h ) - f ( a)$$ doesn't necessarily change sign as $$h$$ approaches $$0$$ from the left and from the right. ... ...
Hope someone can help ...

Peter

 

[HOI]

Yes, |h| doesn't change sign but h does and so using |h| is a mistake. As the author says "this won't work even in dimension 1". Take the very simple case of f(x)= 2x+ 3 at x= 1. Since this is linear, its derivative at any x is the slope, 2. The usual definition of the derivative gives \lim_{h\to 0}\frac{f(1+ h)- f(1)}{h}= \lim_{h\to 0}\frac{(2(1+ h)+ 3)- 5}{h}= \lim_{h\to 0}\frac{2h}{h}= 2. But if we use "|h|" in the denominator instead of "h" that limit does not exist!

If h> 0 then \lim_{h\to 0}\frac{2h}{|h|}=\lim_{h\to 0^+}\frac{2h}{h}= 2 but if h< 0, |h|= -h so \lim_{h\to 0^-}\frac{2h}{|h|}= \lim_{h\to 0}\frac{2h}{-h}= -2. The two onesided limits are not the same so the limit itself does not exist.