Confusion about bounds for $\sin(x) + \cos^3(x)$ I'm working on a limit that requires the squeeze theorem
$$\lim_{x \to -\infty} \frac{x^2(\sin(x)+\cos^3(x))}{(x^2+1) (x-3)}$$
For the bounds, the solution stated the following:
First, note that
$$-1 \leq \sin(x) \leq 1$$
And
$$-1 \leq \cos(x) \leq 1$$
So that
$$-1 \leq \cos^3(x) \leq 1$$
Thus
$$-2 \leq \sin(x) + \cos^3(x) \leq 2$$
While it is technically true that the sum of the outputs of  $\sin(x)$ and $\cos^3(x)$ will always be greater than $-2$ and less than $2$, isn't this false in the sense that the range of  $\sin(x) + \cos^3(x)$ is not $[-2, 2]$?
What I'm driving at is that the range of possible outputs of function $\sin(x) + \cos^3(x)$ does not oscillate between $[-2, 2]$ as the inequality suggests.
Just thinking through this with the simpler function $\sin(x) +\cos(x)$, it would not make sense for the range to be $[-2, 2]$ for it suggests that there are $x$-values for which both sine and cosine are $-1$ or $1$ simultaneously, which we know is false. And since $\cos^3(x)$ is a modification of $\cos(x)$ the same line of reasoning holds.
Graphing $\sin(x) + \cos^3(x)$ we see the actual range is $[-1.172, 1.172]$

My question is why is it valid to use  $-2$ and $2$ as the bounds of $\sin(x) + \cos^3(x)$ despite not being the actual rannge values? Doesn't it falsely suggest that $\sin(x) + \cos^3(x)$ oscillates between $[-2, 2]$?
Problem can be found here (it's problem #7)
And the solution can be found here
 A: We don't need to find the exact upper and lower bounds to obtain the result indeed in this case it suffices to observe that $|\sin(x)+\cos^3(x)|\le 100$ and then
$$-\frac{100x^2}{(x^2+1) (3-x)}\le \frac{x^2(\sin(x)+\cos^3(x))}{(x^2+1) (x-3)}\le \frac{100x^2}{(x^2+1) (3-x)}$$
and apply squeeze theorem with
$$\pm \frac{100x^2}{(x^2+1) (3-x)} \to 0$$
A: An excellent answer has been provided above using squeeze theorem.
Note that
$$ \lim_{x \to -\infty} \frac{x^2(\sin(x)+\cos^3(x))}{(x^2+1) (x-3)}=\lim_{x \to -\infty} \frac{\sin(x)+\cos^3(x)}{(1+\frac{1}{x^2}) (x-3)} =\frac{C}{-\infty}\to 0 $$
A: From the given solution, you should understand that what matters is that this factor is bounded. As the limit is zero anyway, the exact values of the bounds are irrelevant.
So one could have used the gross bracketing
$$0\le|\sin(x)+\cos^3(x)|\le|\sin(x)|+|\cos^3(x)|\le1+1$$
or just said "the factor is obviously bounded".

For tight bounds, find the extrema from
$$\cos(x)-3\sin(x)\cos^2(x)=\cos(x)\left(1-\frac32\sin(2x)\right)=0$$
and
$$\cos(x)=0\to\sin(x)+\cos^3(x)=\pm1,$$
$$\sin(2x)=\frac23\to\sin(x)+\cos^3(x)=\pm\sqrt{\frac{3-\sqrt5}6}\pm\left(\sqrt{\frac{3+\sqrt5}6}\right)^3\\\approx\pm1.1720537521447$$
