# Differentials in R^n ... Another Remark by Browder, Section 8.2 ... ...

#### Peter

##### Well-known member
MHB Site Helper
I am reading Andrew Browder's book: "Mathematical Analysis: An Introduction" ... ...

I am currently reading Chapter 8: Differentiable Maps and am specifically focused on Section 8.2 Differentials ... ...

I need some further help in fully understanding some remarks by Browder made after Definition 8.9 ...

Definition 8.9 and the following remark read as follows:

In the above Remark by Browder we read the following:

"for any fixed $$\displaystyle k \neq 0$$ and $$\displaystyle t \gt 0$$, we have

$$\displaystyle \frac{1}{ |tk| }( L(tk) - M(tk) ) = \frac{1}{|k|},(Lk - Mk )$$ ... ... ... "

My questions are as follows:

Question 1

Browder puts $$\displaystyle h = tk$$ and then lets $$\displaystyle t \to 0$$ ... why is Browder doing this ... what is the logic behind this ... what do we gain by putting $$\displaystyle h = tk$$ ... both $$\displaystyle h$$ and $$\displaystyle k \in \mathbb{R}^n$$ and also isn't $$\displaystyle h$$ just as arbitrary as $$\displaystyle k$$ ... ?

Question 2

How exactly (and in detail) does letting $$\displaystyle t \to 0$$ allow us to conclude that $$\displaystyle Lk = Mk$$ ...

Help will be much appreciated ...

Peter

#### Opalg

##### MHB Oldtimer
Staff member
In the above Remark by Browder we read the following:

"for any fixed $$\displaystyle k \neq 0$$ and $$\displaystyle t \gt 0$$, we have

$$\displaystyle \frac{1}{ |tk| }( L(tk) - M(tk) ) = \frac{1}{|k|},(Lk - Mk )$$ ... ... ... "

My questions are as follows:

Question 1

Browder puts $$\displaystyle h = tk$$ and then lets $$\displaystyle t \to 0$$ ... why is Browder doing this ... what is the logic behind this ... what do we gain by putting $$\displaystyle h = tk$$ ... both $$\displaystyle h$$ and $$\displaystyle k \in \mathbb{R}^n$$ and also isn't $$\displaystyle h$$ just as arbitrary as $$\displaystyle k$$ ... ?

Question 2

How exactly (and in detail) does letting $$\displaystyle t \to 0$$ allow us to conclude that $$\displaystyle Lk = Mk$$ ...
You need to make a clear distinction between fixed and variable vectors. In the equation $$\displaystyle \lim_{h\to0}\frac1{|h|}(Lh - Mh) = 0$$, $h$ has to be a variable vector that converges to $0$. But $k$ is a fixed (nonzero) vector. if $h=tk$ then $h$ varies as $t$ varies. And as $t$ goes to $0$, $h$ also goes to $0$. So you can conclude that $$\displaystyle \lim_{t\to0}\frac1{|tk|}(L(tk) - M(tk)) = 0$$. But $$\displaystyle \frac1{|tk|}(L(tk) - M(tk)) = \frac1{|k|}(Lk-Mk)$$, which is constant. It follows that$$\displaystyle \frac1{|k|}(Lk-Mk) = 0$$, and since $|k|$ is nonzero the conclusion is that $Lk-Mk = 0$.

#### Peter

##### Well-known member
MHB Site Helper
You need to make a clear distinction between fixed and variable vectors. In the equation $$\displaystyle \lim_{h\to0}\frac1{|h|}(Lh - Mh) = 0$$, $h$ has to be a variable vector that converges to $0$. But $k$ is a fixed (nonzero) vector. if $h=tk$ then $h$ varies as $t$ varies. And as $t$ goes to $0$, $h$ also goes to $0$. So you can conclude that $$\displaystyle \lim_{t\to0}\frac1{|tk|}(L(tk) - M(tk)) = 0$$. But $$\displaystyle \frac1{|tk|}(L(tk) - M(tk)) = \frac1{|k|}(Lk-Mk)$$, which is constant. It follows that$$\displaystyle \frac1{|k|}(Lk-Mk) = 0$$, and since $|k|$ is nonzero the conclusion is that $Lk-Mk = 0$.

Thanks Opalg ...