Let $f:(0, \infty) \rightarrow \mathbb{R}$ be a function satisfying$$\lim _{x \rightarrow \infty} \frac{f(x)}{x^k}=a \in \overline{\mathbb{R}}, \quad k \in \mathbb{N}^* \backslash\{1\}$$and$$\lim _{x \rightarrow \infty} \frac{f(x+1)-f(x)}{x^{k-1}}=b \in \overline{\mathbb{R}}$$Prove that $b=k a$.
I have found the following Solution given in the book.
Our assumption implies $\lim _{n \rightarrow \infty} f(n+1)-f(n) / n^{k-1}=b$. Define $u_n=f(n)$ and $v_n=n^k$. Hence$$\begin{aligned}\lim _{n \rightarrow \infty} \frac{u_{n+1}-u_n}{v_{n+1}-v_n} & =\lim _{n \rightarrow \infty} \frac{u_{n+1}-u_n}{n^{k-1}} \cdot \lim _{n \rightarrow \infty} \frac{n^{k-1}}{v_{n+1}-v_n} \\& =\lim _{n \rightarrow \infty} \frac{f(n+1)-f(n)}{n^{k-1}} \cdot \frac{n^{k-1}}{(n+1)^k-n^k}=\frac{a}{k}\end{aligned}$$On the other hand, by the Stolz-Cesáro lemma,$$\lim _{n \rightarrow \infty} \frac{f(n+1)-f(n)}{(n+1)^k-n^k}=\lim _{n \rightarrow \infty} f(n) n^k$$Combining these relations, we deduce that $b=ka$.
My doubt: In the original problem they have given the limit on $x$ which ranges over all reals, however in the solution they have considered limit over a specific real subsequence, more specifically the integers, and concluded results based on that, which doesn't seem justified to me. I do get that both $x$, and $n$ tends to infinity, but in the solution they still consider a specific subsequence of reals, and not all subsequences.
For example in the following function defined from real to real, $f(x)= x \sin(1/x)$, and consider $\lim_{x \to \infty} f(x)$ and then look at the limit on the following subsequence: $a_n =n \pi$ which goes to $\infty$ and $n$ goes to $\infty$ then limit is $0$, which is clearly not the case.