Currently I'm studying limits by definition and I've seen lots of cases where at some point of the proof, for some given function that involves something of the form:
$\sin [f(x_1 \dots x_n)]$ where $f$ is some scalar field
People use the fact that
$\sin [f(x_1 \dots x_n)] \leq f(x_1 \dots x_n)$
Instead of using:
$\sin [f(x_1 \dots x_n)] \leq 1$
Is it bad to always use the latter? Am I loosing some generality or something?
The mean-value theorem gives us the estimate $|\sin x|\leq |x|$ for all $x\in\Bbb{R}$. We also have $|\sin x|\leq 1$ for all $x\in\Bbb{R}$, and therefore, we have for all $x\in\Bbb{R}$, \begin{align} |\sin x|\leq \min(1,|x|). \end{align} Now, sometimes, all we care about is that $\sin$ is a bounded function, in which case we just use the upper bound $1$. On the other hand, sometimes, we may be analyzing a function $f$ which is quite small, i.e $|f|\leq 1$ and in this case if we require more precise estimates on $\sin \circ f$, then we use the second type of estimate. It is all a matter of context.
As an example of a situation where the crude and obvious bound of $1$ is insufficient, consider the following problem:
If you use the crude estimate $|\sin (x^2)|\leq 1$, then you only obtain $\left|\frac{\sin (x^2)}{x}\right|\leq \frac{1}{|x|}$, and this is a useless bound because $\frac{1}{|x|}\to \infty$ as $x\to 0$.
On the other hand, if you use the stronger estimate, then $\left|\frac{\sin(x^2)}{x}\right|\leq \left|\frac{x^2}{x}\right|=|x|$, and now the RHS approaches $0$ as $x\to 0$. SO, in this manner, we are able to prove the limit.