Solving Kantorovitz' Example 3 on Real Valued Functions of Several Variables

[Math Amateur]

I am reading the book "Several Real Variables" by Shmuel Kantorovitz ... ...

I am currently focused on Chapter 2: Derivation ... ...

I need help with an aspect of Kantorovitz's Example 3 on pages 65-66 ...

Kantorovitz's Example 3 on pages 65-66 reads as follows:
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In the above example, we read the following:"... ... \(\displaystyle \frac{ \mid \phi_0 (h) \mid }{ \| h \| } = \frac{ \mid h_1 \text{ sin } (h_2 h_3) \mid }{ \| h \|^{ a + 1 } }\) ... ... ... "
My question is as follows:In the Section on The Differential (see scanned text below) ...

Kantorovitz defines \(\displaystyle \phi_x(h)\) as follows:

\(\displaystyle \phi_x(h) := f(x +h) - f(x) - Lh\)

so that

\(\displaystyle \phi_0(h) := f(0 +h) - f(0 ) - Lh = f(h) - f(0)\) ...... BUT in the Example ... as I understand it ... \(\displaystyle f(0)\) does not exist ...? ...

... BUT ... Kantorovitz effectively gives \(\displaystyle \mid \phi_0 (h) \mid = \frac{ \mid h_1 \text{ sin } (h_2 h_3) \mid }{ \| h \| } \)
Can someone please explain how Kantorovitz gets this value when \(\displaystyle f(0)\) does not exist?Help will be much appreciated ...

Peter============================================================================================

***NOTE***

Readers of the above post may be helped by having access to Kantorovitz' Section on "The Differential" ... so I am providing the same ... as follows:
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[GJA]

Hi, Peter.

Can someone please explain how Kantorovitz gets this value when \(\displaystyle f(0)\) does not exist?

The function is defined by the author to be zero at zero.

This example is akin to the single-variable function $f(x)=\sin(x)/x$, which, in itself, is not defined at zero. However, by using L'Hopital's rule (or a power series expansion), we can see that the limit of this function as $x$ approaches 0 is 1. Hence, we can extend the original function by defining $f(0)=1.$ Doing so produces a continuous (in fact differentiable) function at $x=0$. Looking up examples of "removable discontinuities" online could prove helpful.

 

[Math Amateur]

Hi, Peter.
The function is defined by the author to be zero at zero.

This example is akin to the single-variable function $f(x)=\sin(x)/x$, which, in itself, is not defined at zero. However, by using L'Hopital's rule (or a power series expansion), we can see that the limit of this function as $x$ approaches 0 is 1. Hence, we can extend the original function by defining $f(0)=1.$ Doing so produces a continuous (in fact differentiable) function at $x=0$. Looking up examples of "removable discontinuities" online could prove helpful.

Thanks GJA ...

You write:

" ... ... The function is defined by the author to be zero at zero. ... ... "

Oh! Now how did I miss that ... ! ... :( ...

Thanks for further advice ... most helpful!

Peter