# 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.

$$\frac{1}{2}\leq\frac{x^2+1}{x^2+2}<\frac{x^2+2}{x^2+2}=1.$$ 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 $$\sup_{x\in\mathbb R}\frac{x^2+1}{x^2+2}=1.$$

• Thank you. It is enough? Do I not need to prove anything like I tried for infimum? Jan 12, 2020 at 12:47
• @Uhans It's enough. Maybe also to say that $f(x)=\frac{x^2+1}{x^2+2}$ is a continuous function, but it's not necessary. Jan 12, 2020 at 13:10

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.

$$f(y)= 1- \dfrac{1}{y+2} \le b <1$$, $$y \ge 0$$.

$$0<1-b\le \dfrac{1}{y+2}$$;

$$y+2 \le \dfrac{1}{1-b}$$, $$y \ge 0$$;

Hence $$\sup f(y)=1$$.