$f(x)=f(x+1)$ for all x

Question

Let $f$ be a function defined from $(0,\infty)$ to $\mathbb{R}$

such that

$ \lim_{x \rightarrow \infty } f(x) = 1$ & $f(x)=f(x+1)$ for all $x \in \mathbb{R}$

Then what can you say about $f$

A) continuous and bounded

B) bounded but not necessarily continuous

C) neither necessarily continuous nor necessarily bounded

D) continuous but not necessarily bounded

The question doesnt says any thing about continuity... but i think it is bounded not sure about the answer

Answer 1

Imagine $f(2) = 2$. Then by induction, $f(100000) = 2$. Clearly, the limit can never be $1$.

So, the only way for the limit to be $1$ is if $f(x) = 1$ for all $x$. This is more than enough to determine the function is continuous and bounded.

Answer 2

Let $a\in(0,1]$ and let $x_n = a + n,\ n\geq 0$. Since $\lim_{x\to \infty} f(x) = 1$, it means that $\lim_n f(x_n) = 1$. On the other hand $f(x_n) = f(a)$, so $f(a) = 1$ for all $a\in (0,1]$. Now it follows that $f(x) = 1$, for all $x>0$.