<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 for a more direct approach not involving limits (at least not explicitly): if $x$ is large enough, then you can conclude
\begin{align*}
3x & < (0.01) x^2; \\
1 & < (0.01) x^2; \\
x^2 & < (0.01) x^3; \mathrm{~and} \\
1 & < (0.01) x^3.
\end{align*}
For such $x$, you will have:
$$5x^2 + 3x + 1 < 5x^2 + (0.01) x^2 + (0.01) x^2 = (5.02) x^2$$
and
$$x^3 - x^2 - 1 > x^3 - 0.01 x^3 - 0.01 x^3 = (0.98) x^3.$$
Therefore, you will be able to conclude:
$$\frac{5x^2 + 3x + 1} {x^3 - x^2 - 1} < \frac{(5.02) x^2}{(0.98) x^3} = \frac{5.02}{0.98} \cdot \frac{1}{x} < \frac{10}{x}.$$
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