Differentials in Multivariable Functions .... Kantorovitz: Example 4, page 66

[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 4 on page 66 ...

Kantorovitz's Example 4 on page 66 reads as follows:View attachment 7817In the above example, Kantorovitz show that$$\phi_0 (h) = - \frac{ \| h \|^2 }{( 1 + \sqrt{ 1 + \| h \|^2 )}^2 }$$Kantorovitz then declares that $$\frac{ \phi_0 (h) }{ \| h \| } \rightarrow 0$$ as $$h \rightarrow 0$$ ... ...Can someone please show me how to demonstrate rigorously that this limit is as stated i.e that is that $$\frac{ \phi_0 (h) }{ \| h \| } \rightarrow 0$$ as $$h \rightarrow 0$$ ... ...
... ... 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:View attachment 7818
View attachment 7819
https://www.physicsforums.com/attachments/7820

 

[Ackbach]

In the above example, Kantorovitz show that
$$\phi_0 (h) = - \frac{ \| h \|^2 }{\left( 1 + \sqrt{ 1 + \| h \|^2 }\right)^2 }$$
Kantorovitz then declares that $$\frac{ \phi_0 (h) }{ \| h \| } \rightarrow 0$$ as $$h \rightarrow 0$$ ... ...

Can someone please show me how to demonstrate rigorously that this limit is as stated i.e that is that $$\frac{ \phi_0 (h) }{ \| h \| } \rightarrow 0$$ as $$h \rightarrow 0$$

Well, I don't know about rigorous, but intuitively if you're talking about the limit as $h\to 0$, then $h\not=0$, which forces $\|h\|\not=0$ (in most spaces, at least). Then
$$\phi_0 (h) = - \frac{ \| h \|^2 }{\left( 1 + \sqrt{ 1 + \| h \|^2 }\right)^2 } \; \implies \;
\frac{\phi_0 (h)}{\|h\|}=- \frac{ \| h \|}{\left( 1 + \sqrt{ 1 + \| h \|^2 }\right)^2 }.$$
The denominator is always strictly greater than $4$ (in particular, it's bounded away from zero), and the numerator goes to zero.

 

[S.G. Janssens]

In the above example, Kantorovitz show that$$\phi_0 (h) = - \frac{ \| h \|^2 }{( 1 + \sqrt{ 1 + \| h \|^2 )}^2 }$$

It seems to me that in your book there is a factor $2$ missing in the denominator of $\phi_0(h)$. (The error occurs in the third equality in his example.) So, I think it should be
\[
\phi_0 (h) = - \frac{ \| h \|^2 }{2\left( 1 + \sqrt{ 1 + \| h \|^2}\right)^2 },
\]
but this is innocent: It does not invalidate Ackbach's argument.

Well, I don't know about rigorous, but intuitively if you're talking about the limit as $h\to 0$, then $h\not=0$, which forces $\|h\|\not=0$ (in most spaces, at least). Then
$$\phi_0 (h) = - \frac{ \| h \|^2 }{\left( 1 + \sqrt{ 1 + \| h \|^2 }\right)^2 } \; \implies \;
\frac{\phi_0 (h)}{\|h\|}=- \frac{ \| h \|}{\left( 1 + \sqrt{ 1 + \| h \|^2 }\right)^2 }.$$
The denominator is always strictly greater than $4$ (in particular, it's bounded away from zero), and the numerator goes to zero.

In my opinion this is rigorous: I don't think the author of the book expects the reader to prove the limit from the $(\epsilon,\delta)$-definition, although here that is not hard, but it is just too time-consuming. Instead the reader can resort to the quotient rule for limits, exactly for the reasons you state.

 

[Math Amateur]

It seems to me that in your book there is a factor $2$ missing in the denominator of $\phi_0(h)$. (The error occurs in the third equality in his example.) So, I think it should be
\[
\phi_0 (h) = - \frac{ \| h \|^2 }{2\left( 1 + \sqrt{ 1 + \| h \|^2}\right)^2 },
\]
but this is innocent: It does not invalidate Ackbach's argument.
In my opinion this is rigorous: I don't think the author of the book expects the reader to prove the limit from the $(\epsilon,\delta)$-definition, although here that is not hard, but it is just too time-consuming. Instead the reader can resort to the quotient rule for limits, exactly for the reasons you state.

I now understand the above limit ... thanks to Ackbach and Krylov ...

Thanks to you both ...

Peter