I would like to find a more general way to determine limits of this kind:
$$ \lim_{t\to -\infty} \frac{2-t+\sin t}{t + \cos t}=-1 $$
So I just learned that the limit neither $\cos t$ or $\sin t$ exists, but can I think of these limits as e.g. $\lim_{t\to \pm\infty} \sin t \in [-1,1]$?
Following this logic, is it reasonable to neglect the $\sin t$ and $\cos t$ assuming their value must lie in the interval $[-1,1]$ when $t$ goes to $\pm \infty$?
If I am not mistaken this solution ends up being $-1$ because the limit ends up looking like this when neglecting the $\sin t$ and $\cos t$ when $t \to \pm \infty$. $$ \lim_{t \to - \infty} \frac{2-t}{t}=-\frac{\infty}{\infty}= -1 $$
$$ \lim_{t\to -\infty} \frac{2-t+\sin t}{t + \cos t}= \lim_{t\to -\infty} \frac{t(\frac{2}{t}-1+\frac{\sin t}{t})}{t(1 + \frac{\cos t}{t})}=-1 $$ since it is sufficient to observe that:
1)$\frac{2}{t}\to 0$.
2)$\frac{\sin t}{t}$ and $\frac{\cos t}{t}$ go to $0$ since are both of them the product between a function that goes to $0$ ($\frac{1}{t}$) and a bounded function ($\cos t$ and $\sin t$).