Can you use limits to calculate the supremum/infimum of a set? 
Consider a set $V= \{1+\frac{1}{x+1}, x\in\mathbb{R}^+\cup \{0\}\}$, calculate, if they exists, $\sup(V)$, $\inf(V)$, $\max(V)$ and $\min(V)$.

My answer would be $$\sup(V)=\lim_{n\to0}(\frac{1}{x+1})=2\in V$$, $$\inf(V)=\lim_{x\to\infty}(\frac{1}{x+1})=1 \notin V$$
Hence, $\max(V)=2$ and $\min(V)$ does not exist. However, I am wondering if you are allowed to use limits in order to calculate the $\sup$  and $\inf$. Especially considering the fact that I am asked to calculate the $\max, \min$ too. Intuitive it feels that $2\in V, 1\notin V$, but I have no idea how to prove that without estimation. 

We have
$$
f(x) = 1 + \frac{1}{1+x} \\
D = \mathbb{R_+} \cup \{ 0 \} \\
V = \{ f(x) \mid x \in D \} 
$$
Then
$$
2 \in V \iff \exists \xi \in D: f(\xi) = 2
$$
which is the case for $\xi = 0 \in D$.
Further $1 \not\in V$:
$$
1 \in V \iff \\
\exists \xi\in D: 1 = f(\xi) = 1 + \frac{1}{1+\xi} \iff \\
\exists \xi\in D: 0 = \frac{1}{1+\xi}
$$
But $1/y \ne 0$ for $y \in \mathbb{R}\setminus \{ 0 \}$, 
or $1/(1+\xi) \ne 0$ for $\xi \in \mathbb{R}\setminus \{ -1 \} \supseteq D$.
