Suppose $f\colon \mathbb{R} \to \mathbb{R}$ is a continuous function. Let $x\in \mathbb{R}$ and let $(x_n) \subseteq \mathbb{R}$ be a sequence converging to $x$. Let $(y_n)$ be a subsequence of $(x_n)$.
Is the following true?
$$\lim_{n\to \infty} \frac{|f(y_n)-f(x)|}{|y_n-x|}=\lim_{n\to \infty} \frac{|f(x_n)-f(x)|}{|x_n-x|} \, $$ provided $\lim_{n\to \infty} \frac{|f(y_n)-f(x)|}{|y_n-x|}$ exists or is $\infty$.
I think so but can't use algebra of limits because of the denominator.
I am asking this because I am trying to show that if for every sequence $(x_n)$ converging to $x$ there is a subsequence $(y_n)$ such that $\lim_{n\to \infty} \frac{|f(y_n)-f(x)|}{|y_n-x|}=\infty$, then $\liminf_{z\to x}\frac{|f(y)-f(x)|}{|y-x|}=\infty$.
You can consider both these expressions as sequences themselves, i.e. \[a_n = \frac{|f(y_n)-f(x)|}{|y_n - x|},\quad b_n=\frac{|f(x_n)-f(x)|}{|x_n - x|}.\]
Now, it shouldn't be so hard to see that $y_n$ being a subsequence of $x_n$ means that $a_n$ is a subsequence of $b_n$, and exactly the same conclusions apply: if $b_n$ converges, then $a_n$ converges with the same limit. Note that this means that if you can prove both sequences converge, you can find their common limit by calculating the limit of the subsequence. On the other hand, it is possible for $a_n$ to converge even if $b_n$ does not.
There are functions $f$ that are continuous but not differentiable, and those functions will have sequences like $b_n$ that do not converge with subsequences $a_n$ that do.