Equivalent notion of limit: $f((1+n)x + J) \to C \in \mathbb{R}$ (as $n, J \to \infty$) $\implies$ $f(x) \to C$ (as $x \to \infty$)?

Question

Let $f: [0,\infty) \to \mathbb{R}$ be a bounded continuous function.

Suppose that $$f((1+n)x + J) \to C \in \mathbb{R}$$ for $x$ fixed (and positive) as $n \to \infty$ ($n \in \mathbb{N}$) and $J \to \infty$. How can I prove that we have

$$f(x) \to C$$ as $x \to \infty$?

Answer 1

Let's show that $f(y)\rightarrow C$ as $y\rightarrow \infty$ - changing $x$ into $y$ in order to avoid confusion.

For any real $y$:

Define $n(y)=floor({y/2x})-1$ as the biggest integer s.t. $(1+n)x\leqslant y/2$

Define $J(y)=y-(1+n(y))x$

$n(y)$ and $J(y) \rightarrow \infty$ when $y\rightarrow \infty$ (both are equivalent to $y/2$)

Applying your formula, you get that

$f(y)=f((1+n(y))x+J(y))\rightarrow C$ when $y\rightarrow \infty$.