<s>[INST] 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.
 [/INST] $$\frac{1}{2}\leq\frac{x^2+1}{x^2+2}<\frac{x^2+2}{x^2+2}=1.$$
The equality in the right inequality occurs for $x=0$ and since $$\lim_{x\rightarrow+\infty}\frac{x^2+1}{x^2+2}=1,$$ we are done:
$$\inf_{x\in\mathbb R}\frac{x^2+1}{x^2+2}=\frac{1}{2}$$ and
$$\sup_{x\in\mathbb R}\frac{x^2+1}{x^2+2}=1.$$
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