Find supremum and infimum for $C={\frac{x^2+1}{x^2+2}: x \in \mathbb{R}}.$ Find supremum and infimum for
$C={\frac{x^2+1}{x^2+2}: x \in \mathbb{R}}.$
We can easy see that $\frac{x^2+1}{x^2+2}$ is bounded from below with $m=\frac{1}{2}.$ This is also infimum. To show this I proceed like this.
Let us assume that $\frac{1}{2}$ is infinum (is not the largest bound for $\frac{x^2+1}{x^2+2}$), i.e there exists $m>\frac{1}{2}$ such that $\frac{x^2+1}{x^2+2}\geq m$. Let us take $x=0$. We have $\frac{0^2+1}{0^2+2}=\frac{1}{2}\geq m>\frac{1}{2}.$vSince we get $\frac{1}{2}>\frac{1}{2}$ which is not true, so we conclude that $\frac{1}{2}$ is infinum.
For supremum, we can see that $\frac{x^2+1}{x^2+2}=1-\frac{1}{x^2+2}$ is bounded from above with $M=1$ and I think this is supremum (the smallest bound). I do not know how to prove it.vI would be grateful for any help.

Attempt:
$y:=x^2$; $y\ge 0$;
$f(y):=\dfrac{y+1}{y+2}= 1-\dfrac{1}{y+2}$.
1) $\dfrac{1}{y+2} \le 1/2$.
Equality for $y=0$.
$\min f(y)= \inf f(y) =1/2$;
2) $\dfrac{1}{y+2} \gt 0$;
Hence $1$ is an upper bound for $f(y)$.
Assume there is a smaller upper bound $b >0$, i.e.
$f(y)= 1- \dfrac{1}{y+2} \le b <1$, $y \ge 0$.
$0<1-b\le \dfrac{1}{y+2}$;
$y+2 \le \dfrac{1}{1-b}$, $y \ge 0$; 
a contradiction.
Hence $\sup f(y)=1$.
