Proof of Chain Rule: Understanding the Limits

[mcastillo356]

TL;DR

I've got a proof of the Chain Rule, but basic questions about basic steps

First I quote the text, and then the attempts to solve the doubts:

"Proof of the Chain Rule

Be $f$ a differentiable function at the point $u=g(x)$, with $g$ a differentiable function at $x$. Be the function $E(k)$ described this way:
$$E(0)=0$$
$$E(k)=\dfrac{f(u+k)-f(u)}{k}-f'(u)\qquad\mbox{if}\;k\neq 0$$

By definition of derivative, $\lim_{k \to{0}}{E(k)}=f'(u)-f'(u)=0=E(0)$, so $E(k)$ is continuous in $k=0$. Also, be $k=0$ or not, we have

$$f(u+k)-f(u)=(f'(u)+E(k))k$$

Be now $u=g(x)$ and $k=g(x+h)-g(x)$, so $u+k=g(x+h)$; we obtain

$$f(g(x+h))-f(g(x))=(f'(g(x))+E(k))(g(x+h)-g(x))$$

As $g$ is differentiable at $x$, $\lim_{h \to{0}}{[g(x+h)-g(x)]/h}=g'(x)$. Also, $g$ is continuous at $x$, by Theorem 1

Spoiler: Theorem 1

THEOREM 1 Being differentiable means being continous

If $f$ is differentiable at $x$, we know it exists

$$\displaystyle\lim_{h \to{0}}{\dfrac{f(x+h)-f(x)}{h}}=f'(x)$$

Using the rules of limits (Theorem 2 of section 1.2)

Spoiler: Theorem 2 of section 1.2

Rules for limits

If $\lim_{x \to a}{f(x)}=L$, $\lim_{x \to a}{g(x)}=M$, and $k$ is a constant, then

1. Limit of a sum: $\displaystyle\lim_{x \to{a}}{[f(x)+g(x)]}=L+M$
2. Limit of a subtraction: $\displaystyle\lim_{x \to{a}}{[f(x)+g(x)]}=L+M$
3. Limit of a product: $\displaystyle\lim_{x \to{a}}{f(x)g(x)}=LM$
4. Limit of a function multiplied by a constant: $\displaystyle\lim_{x \to{a}}{kf(x)}=kL$
5. Limit of a division: $\displaystyle\lim_{x \to a}{\dfrac{f(x)}{g(x)}}=\dfrac{L}{M}\qquad\mbox{if}\;M\neq 0$
If $m$ is an integer and $n$ a positive integer, then
6. Limit of a power: $\displaystyle\lim_{x \to{a}}{[f(x)]^{m/n}}=L^{m/n}$, whenever $L>0$ if $n$ is even, and $L\neq 0$ if $m<0$
If $f(x)\geq g(x)$ at an interval that contains $a$ inside, then
7. Order preservation: $L\geq M$

we have

$$\lim_{h \to{0}}{f(x+h)-f(x))}=\lim_{h \to{0}}{\left (\dfrac{f(x+h)-f(x)}{h}\right)(h)}=(f'(x))(0)=0$$

This is equivalent to $\lim_{h \to{0}}{f(x+h)=f(x)}$, and means $f$ is continous.So $\lim_{h \to{0}}{E(k)}=\lim_{h \to{0}}{(g(x+h)-g(h)=0}$. As $E$ is continuous in 0, $\lim_{h \to{0}}{E(k)}=\lim_{k \to{0}}{E(k)}=E(0)=0$. This way,

$$\dfrac{d}{dx}f(g(x))=\displaystyle\lim_{h \to{0}}{\dfrac{f(g(x+h))-f(g(x))}{h}}$$

$$ \qquad\qquad\qquad=\displaystyle\lim_{h \to{0}}{(f'(g(x)+E(k))\dfrac{g(x+h)-g(x)}{h}}$$

$$\qquad\qquad\qquad=(f'(g(x)+0)g(x)=f'(g(x))g'(x)$$

As we wanted to prove."

Doubts:

-¿$\lim_{h \to{0}}{E(k)}=\lim_{k \to{0}}{E(k)}$?

Attempt: $\lim_{h \to{0}}{E(k)}=\displaystyle\lim_{h \to{0}}{\dfrac{E(k+h)-E(k)}{h}}=\displaystyle\lim_{k \to{0}}{\dfrac{E(k)-E(0)}{k-0}}$ (Bad, I guess)

-¿Why $f'(g(x))$ remains the same at the last step?:

$$ \qquad\qquad\qquad=\displaystyle\lim_{h \to{0}}{(f'(g(x)+E(k))\dfrac{g(x+h)-g(x)}{h}}$$

$$\qquad\qquad\qquad=(f'(g(x)+0)g(x)=f'(g(x))g'(x)$$

Attempt: are different variables

Greetings to everybody, have a nice St Joseph's Day!

 

[Mark44]

-¿limh→0E(k)=limk→0E(k)?

No. $\lim_{h \to 0} E(k) = E(k)$, since E(k) doesn't involve h in any way.

For the same reason, if f(x) = 2x + 3, $\lim_{z \to 19} f(x) = f(x)$

 

[mcastillo356]

Thanks!
Greetings

 

[Mark44]

-¿Why f′(g(x)) remains the same at the last step?:

For the same reason as before -- f'(g(x)) doesn't involve h, so in the limit process (with h approaching 0), f'(g(x)) is unaffected.