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
Best Answer
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
$$\pm \frac{100x^2}{(x^2+1) (3-x)} \to 0$$