I have some question about an exercise in Abbott's Understanding Analysis.
Prove the following lemma:
Lemma. Let $f$ be defined on an open interval $J$ and assume that $f$ is differentiable at some $a\in J$. If $(a_n)$ and $(b_n)$ are sequences satisfying $a_n<a<b_n$ and $\lim a_n = a = \lim b_n$, then $$f'(a) =\lim_{n\rightarrow\infty}\frac{f(b_n)-f(a_n)}{b_n-a_n}.$$
Here's a possible proof:
$$\begin{align}\lim_{n\rightarrow\infty}\frac{f(b_n)-f(a_n)}{b_n-a_n} &=\lim_{n\rightarrow\infty}\frac{f(b_n) - f(a)}{b_n - a}\cdot \frac{b_n - a}{b_n - a_n}+ \lim_{n\rightarrow\infty} \frac{f(a) - f(a_n)}{a-a_n} \cdot \frac{a-a_n}{b_n - a_n} \tag{1} \\ &=f'(a)\lim_{n\rightarrow\infty}\frac{b_n - a}{b_n - a_n}+f'(a)\lim_{n\rightarrow\infty} \frac{a-a_n}{b_n - a_n} \tag{2} \\ &= f'(a) \lim_{n\rightarrow\infty}\left(\frac{b_n-a+a-a_n}{b_n-a_n}\right) \tag{3} \\ &=f'(a)\end{align}$$
First, I think we require that the sequences $(a_n)$ and $(b_n)$ in the theorem are both contained in $J$. Because according to theorem 4.2.3 in the book, $\lim_{x\rightarrow a}f(x)=L$ is equivalent to $f(x_n)\to L$ for all sequences $(x_n)\to c$ (but $x_n\neq c$ for all $n$) and $(x_n)\subseteq A$, where $A$ is the domain of $f$. This equivalence is used in the above solution (from $(1)$ to $(2)$).
Second, if $\lim a_n$ and $\lim b_n$ exist and are finite, then $\lim (a_n+b_n)=\lim a_n+\lim b_n$ and $\lim (a_nb_n)=\lim a_n\lim b_n$. However, how do we know that this is the case in the above calculation, which seems to be using these laws? Moreover, is it possible to weaken the assumption $a_n<a<b_n$ to say, $a_n\leq a<b_n$?
In the possible proof the first three equalities cannot be justified, because the limits $$\lim\frac{b_n-a}{b_n-a_n}\qquad\text{and}\qquad\lim\frac{a-a_n}{b_n-a_n}\tag{$\ast$}$$ need not exist. It is clear that each sequence in $(\ast)$ is bounded between $0$ and $1$, but that's not enough to conclude that they converge. Hence we cannot apply the laws $\lim(x_n+y_n)=\lim x_n + \lim y_n$ and $\lim x_ny_n=\lim x_n\lim y_n$.
For example define $$a:=0,\qquad a_n:=-\frac1n,\qquad b_n:=\begin{cases}\frac2n&\text{$n$ odd}\\\frac1{2n}&\text{$n$ even}\end{cases}.$$ You can check that these $(a_n)$ and $(b_n)$ satisfy the hypotheses of the lemma, but the sequences in $(\ast)$ alternate between $\frac13$ and $\frac23$ so their limits don't exist.