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I'm struggling with the following problem:

Let $(x_n)$ be an arbitrary sequence of real numbers. Now we want to show that the sequence $y_n = \frac{3x_n + 2}{1+x_n^2}$ is bounded.

I have tried playing around with polynomials and rearranged things for some arbitrary bounds but I couldn't get it to work.

Any help would be appreciated -> Hints are especially welcome.

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    $\begingroup$ Can you obtain upper and lower bounds on the function $f(x)=(3x+2)/(1+x^2)$? $\endgroup$ Apr 19, 2021 at 18:00

2 Answers 2

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A completely elementary argument is possible. Clearly $$|x|^2-2|x|+1=(|x|-1)^2\ge 0$$ for all $x\in\Bbb R$, so $$x^2+1=|x|^2+1\ge 2|x|$$ for all $x\in\Bbb R$, and therefore

$$\left|x+\frac1x\right|=|x|+\frac1{|x|}\ge 2\tag{1}$$

for all $x\in\Bbb R\setminus\{0\}$. (The equality follows from the fact that $x$ and $\frac1x$ have the same sign.)

Let $f(x)=\frac{3x+2}{1+x^2}$. For $x\ne 0$ we have

$$\begin{align*} |f(x)|&=\left|3\left(\frac{x}{1+x^2}\right)+\frac2{1+x^2}\right|\\ &\le3\left|\frac{x}{1+x^2}\right|+\frac2{1+x^2}\\ &\le3\left|\frac{x}{1+x^2}\right|+2\\ &=\frac3{\left|\frac1x+x\right|}+2\\ &\overset{*}\le\frac32+2\\ &=\frac72\,, \end{align*}$$

where the starred step follows from $(1)$, and $f(0)=2$, so $|f(x)|\le\frac72$ for all $x\in\Bbb R$.

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Define$$\begin{array}{rccc}f\colon&\Bbb R&\longrightarrow&\Bbb R\\&x&\mapsto&\displaystyle\frac{3x+2}{1+x^2}.\end{array}$$It is a continuous function and $\lim_{x\to\pm\infty}f(x)=0$. Therefore, $f$ is a bounded function. And if $M\in\Bbb R_+$ is such that$$(\forall x\in\Bbb R):|f(x)|\leqslant M,$$then$$(\forall n\in\Bbb N):|y_n|\leqslant M,$$since $y_n=f(x_n)$.

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