If $\lim\limits_{m\to\infty}\frac{f(a+h_m)-f(a)}{h_m}$ does not exist, with $\{h_m\}\to 0$, why can we conclude $f$ not differentiable at $a$? If we have a function $f(x)$ and we prove that at a point $a$ the limit
$$\lim\limits_{m\to\infty} \frac{f(a+h_m)-f(a)}{h_m}$$ does not exist, where $\{h_m\}$ is a sequence that converges to $0$, why can we conclude that $f$ is not differentiable at $a$?
In other words, why can we conclude something about the differentiability of a function at a point $a$ based on the limit of a sequence?
I think it is related to the following theorem
Spivak, Calculus, Ch. 22, Theorem 1 Let $f$ be a function defined in an open interval containing $c$ except perhaps at $c$. Then

$$\lim\limits_{x\to c} f(x)=l$$
$$\iff$$
for every sequence $\{a_n\}$ such that

*

*each $a_n$ is in the domain of $f$

*each $a_n\neq c$

*$\lim\limits_{n\to\infty} a_n=c$
the sequence $\{f(a_n)\}$ satisfies
$$\lim\limits_{n\to\infty} f(a_n)=l$$

Let $\{a_n\}=\{a+h_m\}$ be any sequence with $\{h_m\}$ converging to $0$. Then

*

*$a_n\in \mathbb{R}$

*$a_n\neq a$

*$\lim\limits_{n\to\infty} a_n=a$
Let $g(x)=\frac{f(x)-f(a)}{x-a}$. Then, $g(a_n)=\frac{f(a+h_m)-f(a)}{h_m}$ is a sequence
but we showed previously that it doesn't converge, ie
$$\lim\limits_{n\to\infty} g(a_n)$$
does not exist.
This seems to mean that the consequent of the theorem above is false. Hence the antecedent is false. And that means that the limit
$$\lim\limits_{x\to a} g(x)=\lim\limits_{x\to a} \frac{f(x)-f(a)}{x-a}$$
does not exist, and therefore, $f$ is not differentiable at $a$.
Is this the underlying justification for our conclusion about differentiability of the function $f$ at $a$ based on a limit of a sequence?

Your argument is fine, but too lengthy.
Suppose $\lim_{h\to 0}g(h)=l$ exists for $g$ defined in a punctured neighborhood $U$ of $0$. Then, for every sequence $(h_m)$ in $U\setminus\{0\}$ such that $\lim_{m\to\infty}h_m=0$, it holds that $\lim_{m\to\infty}g(h_m)=l$.
This is a small part of the big theorem you quote and the easier one. Indeed, choose $\varepsilon>0$. Then there exists $\delta>0$ such that, for $0<|h|<\delta$, $h\in U$, it holds that $|g(h)-l|<\varepsilon$. Since the sequence converges to $0$, there exists $n$ such that, for $m>n$ it holds that $|h_m|<\delta$. Thus, for $n>m$ it holds that $|g(h_m)-l|<\varepsilon$.
Now apply to $g(h)=(f(a+h)-f(a))/h$.
