Introduction. I memorized the problem and I could be getting the constant term wrong in the numerator inside the absolute value, but I feel it won't matter. For instance, I don't think we would have cancelled anything there. My concern is that I don't know how to verify my answer here in any interesting way. How would I verify this?
Problem. Suppose
$$\left|f(x)- \frac{7x^2 + 5x|x| + 2}{x^2 + 16}\right| \leq \frac{1}{x^2}$$
Find $\lim f(x)$ as $x\to \pm\infty$.
A solution. I'll show my approach for $x\to -\infty$. I open the absolute value and rewrite the polynomial to $7x^2 - 5x^2 + 2$ because $|x| = x$ and $x < 0$ when $x \to -\infty$. I then added the rational function from both inequalities and then divided every term in the rational function by $x^2$. I got
$$-\frac{1}{x^2} + \frac{2 + 2/x^2}{1 + 16/x^2} \leq f(x) \leq \frac{2 + 2/x^2}{1 + 16/x^2} + \frac{1}{x^2}$$
Letting $x\to -\infty$, the left term goes to $2$ and the right term goes to $2$ too, so $f(x) \to 2$ as $x\to -\infty$.
Question. How do I verify the approach here? I don't even know how to ask a CAS to verify this.