<s>[INST] Horizontal asymptotes of function defined as an inequation I was given as a homework:
Suppose $f$ is a function of $x$ and, $\forall x\in \mathbb{R}$, it holds
$$
\left| f(x)-\frac{1}{x}\right| \le \frac{2x^2+x|x|+2}{x^2+1}
$$
Find the horizontal asymptotes of $f(x)$.
I tried to use the Squeeze Theorem, rewritting the expression as:
$$
-\frac{2x^2+x|x|+2}{x^2+1} \le  f(x)-\frac{1}{x} \le \frac{2x^2+x|x|+2}{x^2+1}
$$
Which, rearranging, gives
$$
\frac{-2x^3-x^2|x|-2x+x^2+1}{x^3+x} \le  f(x) \le \frac{2x^3+x^2|x|+2x+x^2+1}{x^3+x}\\
g(x) \le  f(x) \le h(x)
$$
The limits, however, are not equal:
$$
\lim_{x\rightarrow\infty}g(x)=-3\\
\lim_{x\rightarrow\infty}h(x)=3\\
\lim_{x\rightarrow-\infty}g(x)=-1\\
\lim_{x\rightarrow-\infty}h(x)=1\\
$$
And I can't use the Squeeze Theorem to find the horizontal asymptotes. Any hints on how to solve it?
 [/INST] There will be no hints on how to solve it because the condition you gave is not enough to completely determine horizontal asymptotes. For $x < 0$, the bounding expression becomes $$\frac{2x^2+x|x|+2}{x^2+1} = \frac {x^2 + 2}{x^2+1} = 1 + \frac 1{x^2+1}$$
which converges to $1$ as $x \to -\infty$, while for $x > 0$, it becomes
$$\frac{2x^2+x|x|+2}{x^2+1} = \frac {3x^2 + 2}{x^2+1} = 3 - \frac 1{x^2+1}$$
which converges to $3$ as $x \to \infty$. Thus choosing $f(x) = \frac1x + a$ will satisfy the inequality for sufficiently large values of $|x|$ for any $a$ with $|a| \le 1$. 
Thus we can find functions $f(x)$ satisfying the condition and having any horizontal asymptote between $-1$ and $1$. Further, there is no reason a function cannot have different horizontal asymptotes to the left and to the right. For example, $\tan^{-1}x$ has a left asymptote of $-\frac \pi 2$ and a right asymptote of $\frac \pi 2$. Your condition requires the left asymptote $f(x)$ to be between $-1$ and $1$, but the right asymptote can be anywhere between $-3$ and $3$.
But that is assuming that the asymptotes exist at all. Your condition is not sufficient to require even that: $f(x) = \frac 1x + a\sin x$ satisfies the inequality for large $x$, but has no horizontal asymptotes.
(I haven't bothered to determine if these functions satisfy the inequality everywhere because it doesn't matter - if it doesn't, we can replace $a$ with a function having $a$ as horizontal asymptote, and dropping to $0$ anywhere necessary to satisfy the inequality.)
</s>