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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.

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2 Answers 2

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$$\frac{1}{2}\leq\frac{x^2+1}{x^2+2}<\frac{x^2+2}{x^2+2}=1.$$ 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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  • $\begingroup$ Thank you. It is enough? Do I not need to prove anything like I tried for infimum? $\endgroup$
    – Uhans
    Jan 12, 2020 at 12:47
  • $\begingroup$ @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. $\endgroup$ Jan 12, 2020 at 13:10
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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 $\sup f(y)=1$.

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