In Chapter 1: Functions and limits, 1.7 The Precise Definition of a Limit,
Let $f$ be a function ... the limit of $f(x)$ as $x$ approaches $a$ is $L$, and we write $$\lim_{x\to a }f(x)=L$$ if for every number $\epsilon>0$ there is a number $\delta>0$ such that if $0<|x-a|<\delta$ then $|f(x)-L|<\epsilon$.
OK, let's assume this is the case and that this definition is perfectly true, but:
- Then how do you go about determining this $L$ by the definition? For example if I wanted to compute $\lim_{x\to0}\frac{\sin x}{x}$ how can I determine this $L$ from the definition (without pre-knowledge of the limit)? (this is related to the question below)
- Can we find a suitable $\delta_1$ such that $|f(x)-L_1|<\epsilon$ and a different $\delta_2$ such that $|f(x)-L_2|<\epsilon$? If so then what is right $L_1$ or $L_2$?
I think if those issues are cleared to me then the rest will also be clear.
Thanks in advance.
The definition doesn't show you how to determine $L$. There is no general way to determine $L$. This definition only shows that at most one $L$ exists (see below.) If there was a general way to determine $L$, calculus would be a nothing-burger.
You can prove from this definition that there is only one limit, at most.
For any particular $\epsilon$ you can find two $L_1$ and $L_2$ and $\delta_1$ and $\delta_2$, but there is no way for it to work for all $\epsilon.$ The definition says that for all $\epsilon>0$. This is provable from the definition:
If $L_1\neq L_2$ let $\epsilon=|L_1-L_2|/2$. Try to pick $\delta_1$ and $\delta_2$ for this $\epsilon$. One can prove that this is not possible.