Proving $\lim_{x \to c}f(g(x)) = f( \lim_{x \to c} g(x))=f(L) \text{ where } \lim_{x \to c} g(x) = L \text{ and } f(x) \text{ is continuous at } x=L$

Question

I am new to the $\epsilon - \delta$ definition of a limit and I'm trying to understand this proof.
Prove that $\displaystyle \lim_{x \to c}f(g(x)) = f( \lim_{x \to c} g(x))=f(L) \text{ where } \lim_{x \to c} g(x) = L \text{ and } f(x) \text{ is continuous at } x=L$
This is the proof:
$\displaystyle \lim_{u \to L} f(u) = f(L) \text{ means } \forall \epsilon>0 \hspace{0.2cm} \exists \delta_1 > 0 \text{ such that } 0<|u-L|<\delta_1 \implies |f(u)-f(L)|<\epsilon$
$\displaystyle \lim_{x \to c} g(x) = L \text{ means } \forall \epsilon_1>0 \hspace{0.2cm} \exists \delta > 0 \text{ such that } 0<|x-c|<\delta \implies |g(x)-L|<\epsilon_1$
Then here he substitutes $\delta_1$ instead of $\epsilon_1$, so the statement becomes this:
$\displaystyle \forall \delta_1>0 \hspace{0.2cm} \exists \delta > 0 \text{ such that } 0<|x-c|<\delta \implies |g(x)-L|<\delta_1$
I understand the fact that because $\epsilon_1$ is any number we choose, we can just plug in $\delta_1$, but what I don't understand is why the statement above is true $\forall \delta_1>0$. What if we choose a $\delta_1$ value that doesn't have an $\epsilon$ value associated with it. We know that $\forall \epsilon>0 \hspace{0.2cm} \exists \delta_1>0 \text{ but this doesn't imply that } \forall \delta_1>0 \hspace{0.2cm} \exists \epsilon>0$. Here is the rest of the proof:
let $u=g(x)$, so the statements become this:
$\displaystyle \forall \delta_1>0 \hspace{0.2cm} \exists \delta > 0 \text{ such that } 0<|x-c|<\delta \implies |g(x)-L|<\delta_1$ $\displaystyle\forall \epsilon>0 \hspace{0.2cm} \exists \delta_1 > 0 \text{ such that } 0<|g(x)-L|<\delta_1 \implies |f(g(x))-f(L)|<\epsilon$
$\displaystyle \therefore\forall \delta_1>0 \hspace{0.2cm} \exists \delta > 0 \text{ such that } 0<|x-c|<\delta \implies |f(g(x))-f(L)|<\epsilon$
Then he just checks $g(x) = L \implies |f(g(x))-f(L)|=0<\epsilon$
Thank you all in advance.

Answer 1

This is notation heavy, so it's easy to get confused. You can just plug in $\delta_1$ for $\epsilon_1$ because $\epsilon_1$ is arbitrary, as you say. $\delta_1$ is just a number. We don't know or care whether or not 'there is an $\epsilon$ associated to it'. The author is, perhaps confusingly, setting up the proof in a particular way.

I might 'rewrite' it as follows. Fix an $\epsilon>0$. Because $f$ is continuous, I know that there is a $\delta'>0$ with $|L-y|<\delta$ implying $|f(y)-f(L)|<\epsilon$, (among all $y$ in the domain of $f$).

Because $\lim_{x\to c}g(x)=L$, I know that there is a $\delta>0$ such that $0<|x-c|<\delta$ implies $|g(x)-L|<\delta'$ (among all $x$ in the domain of $g$), where $\delta$ is the same number I used above (everything is fixed).

The chain of dependencies: $\epsilon$ is 'free', $\delta'$ depends on $\epsilon$, and $\delta$ depends on $\delta'$.

Now let $x$ be any number with $0<|x-c|<\delta$. I want to check, is $|f(g(x))-f(L)|<\epsilon$? If so, then the proof is done, because $x$ is arbitrary, this value of $\delta$ 'works, and $\epsilon$ is arbitrary.

I know $|g(x)-L|<\delta'$. That means $y=g(x)$ is a number with $|y-L|<\delta'$, so by definition of $\delta'$ we know $|f(y)-f(L)|<\epsilon$. That is, $|f(g(x))-f(L)|<\epsilon$. We are done.

Answer 2

For any $\epsilon_1 > 0$ there exists a $\delta_1 > 0$ such that $|x-c|<\delta_1 \implies |g(x) - L|<\epsilon_1$

And, for any $\epsilon_2 > 0$ there exists a $\delta_2 > 0$ such that $|u- g(L)|<\delta_2 \implies |f(u) - f(g(L))|<\epsilon_2$

In the first condition, we can choose ANY value for $\epsilon_1.$ In the second condition, we know that $\delta_2$ exists.

That means we can set $\epsilon_1 = \delta_2$

Answer 3

$\displaystyle \lim_{u \to L} f(u) = f(L) \text{ means } \forall \epsilon>0 \hspace{0.2cm} \exists \delta_1 > 0 \text{ such that } 0<|u-L|<\delta_1 \implies |f(u)-f(L)|<\epsilon$
assume that you were given an arbitrary $\displaystyle \epsilon_1$ then you have an $\displaystyle \delta_1$ in which the proposition above is true. since $\displaystyle \delta_1>0$ we can simply let our choice of $\displaystyle \epsilon_1 $ be equle to $\displaystyle \delta_1$ since $\displaystyle \epsilon_1 =\delta_1 > 0 $ and it is guaranteed that for every $\displaystyle \epsilon_1 >0 $ there is $\displaystyle \delta$ such that proposition : $\displaystyle \lim_{x \to c} g(x) = L \text{ means } \forall \epsilon_1>0 \hspace{0.2cm} \exists \delta > 0 \text{ such that } 0<|x-c|<\delta \implies |g(x)-L|<\epsilon_1$ holds

then it is also the case that for every $\displaystyle \epsilon_1 =\delta_1 > 0 $ there exists $\displaystyle \delta$ such that : $\displaystyle \ 0<|x-c|<\delta \implies |g(x)-L|<\delta_1$ hence the proposition : $\displaystyle \forall \delta_1>0 \hspace{0.2cm} \exists \delta > 0 \text{ such that } 0<|x-c|<\delta \implies |g(x)-L|<\delta_1$ is true as we have argued about it.