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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?

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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.)

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