In the proof of the product law of limits, (by the ε-δ definition, I mean) you assume:
$$\lim_{x \rightarrow a} f(x)= L \qquad \lim_{x \rightarrow a} g(x) = M$$ and want to prove that
$$\lim_{x \rightarrow a} (f.g)(x)= L.M$$ By the definition, we got to find an $δ$ such that given an $ε>0$,
$$| x-a | < δ \Rightarrow | f(x).g(x)-L.M | < \varepsilon$$
After some manipulation, we get :
$$ | f(x).g(x)-L.M | < | f(x) | . | (g(x)-M) | + | M | . | f(x)-L | $$
Then, in the proof, we usually let $δ_1=1$ and proceed from there.
My question is: Why do we have to assume first that $δ_1=1$?
My suggestion would be:
Let, for some $δ_1, | f(x)-L | < \frac{ \varepsilon }{2. | M | } $
Then, we would get
$$| f(x).g(x)-L.M | < | f(x) | . | (g(x)-M) | + \frac{ \varepsilon }{2}$$
$$| f(x)-L | < \frac{ \varepsilon }{2. | M | }$$
Because of this assumption,
$$| f(x)-L | < \frac{ \varepsilon }{2. | M | } \Rightarrow | f(x) | < \frac{ \varepsilon }{2. | M | } + | L |$$
Now we got
$$| f(x).g(x)-L.M | < ( \frac{ \varepsilon }{2. | M | } + | L | ) . | (g(x)-M) | + \frac{ \varepsilon }{2} $$
If we just let, for some $δ_2, | (g(x)-M) | < \frac{ \frac{ \varepsilon }{2} }{( \frac{ \varepsilon }{2. | M | } + | L | ) }$, wouldn't that let us say that
$(| f(x).(g(x)-M| + | M | . | f(x)-L |) < \frac{ \varepsilon }{2} + \frac{ \varepsilon }{2} = \varepsilon$? (choosing the minimum between the $2$ deltas)
that means that your delta restrictions would have some $x'$s on them. Is that a problem?
Thank you so much!