<s>[INST] Justifying $\frac{5x^2 + 3x + 1}{x^3 - x^2 - 1} < \frac{10}{x}$ for large $x$ 
I need to find a way to justify the inequality $$\frac{5x^2 + 3x + 1}{x^3 - x^2 - 1} < \frac{10}{x}$$ that holds for large values of x.

As an instance, I will justify an example inequality just to show you the intended strategy. Consider the inequality
$$\left|\frac{5x + 21}{2x^2 - 7}\right| < \frac{5}{x}.$$ For example, $2x^2 - 7$ can be written as $x^2 + (x^2 - 7)$. Because $x^2 - 7$ is a positive value for all $x < -\sqrt{7}$, removing it from the denominator makes the absolute value of the fraction greater. Also note that when $x < -\frac{21}{10}$, the numerator $|5x + 21| < 5|x|$, and this happens for all $x < -\sqrt{7}$. So $$\left|\frac{5x + 21}{2x^2 - 7}\right| < \frac{5|x|}{x^2} = \frac{5}{x}$$ as long as $x < -\sqrt{7}$.
 [/INST] Hint:
Set
$$
f(x)=\frac{5x^3 + 3x^2 + x}{x^3 - x^2 - 1}.
$$
Then you have
$$
\lim_{x\to+\infty}f(x)=5\tag{1}
$$
Now use the limit definition to show that $f(x)<10$ for sufficiently large $x$. So you get
$$
\frac{f(x)}{x}<\frac{10}{x}
$$
for sufficiently large $x$.

Notes: The limit (1) implies that
$$
|f(x)-5|<1
$$
for sufficiently large $x$. So you have
$$
4<f(x)<6<10.
$$
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