Hello I am currently trying to prove a result and I have basically whittled it down to showing the following is true.
Let $I\subset\mathbb{R}$ be an interval and fix $\alpha>1$ real number. Fix $x\in I$ and suppose $f:I \rightarrow \mathbb{R}$ is a strictly increasing function. Let $\{x_k\}$ be a sequence contained in $I$ such that $x_k \neq x$ for all k and $x_k \rightarrow x$. Now suppose that the limit $\lim_{k \to \infty} \frac{f(x_k)-f(x)}{f(x+\alpha(x_k-x))-f(x)} $ exists and is equal to $l$. Then $l \neq 0$.
To me it makes sense that this is true. Since $f$ is strictly increasing it can't do anything too mental near $x$, since $f$ is locally bounded both must converge to 0 and surely the numerator can't dominate the denominator enough for the limit to be zero. My argument in trying to prove it so far goes:
It is clear that $0\leq l \leq 1$ so suppose $l=0$. Then since $f$ is locally bounded we deduce that $f(x_k)-f(x) \to 0$. Since $f$ is strictly increasing we deduce that $f$ must be continuous (here we either get full continuity or left continuity or right continuity depending on the sequence $x_k$ but I don't think this really matters much) at $x$. Then I've been playing around with functions and this seems to hold for everything but I don't really know how to proceed.
Let $I=(-1,1)$ and $$ f(x)=\begin{cases} e^{-1/x} & \text{if }x>0,\\0 & \text{if }x=0, \\ -e^{1/x} & \text{if }x<0. \end{cases}$$ Let $x=0$ and $x_k=1/k$. Then $x+\alpha(x_k-x)=\alpha/k$ and $$ \frac{f(x_k)-f(x)}{f(x+\alpha(x_k-x))-f(x)}=\frac{e^{-k}}{e^{-k/\alpha}}=e^{k(1/\alpha-1)}, $$ which converges to $0$ as $k\to\infty$ since $1/\alpha-1<0$.