Is it possible to solve $lim_{xto a}frac{sin x - sin a }{x-a}$ without derivatives?

My teacher replaced $x-a = t$ and then said as $x$ approaches $a$ we have
$a-a=t$ so $t$ approaches $0$
and then said lim as $t$ approaches $0$, $\frac{\sin(t+a)-\sin a}t = \lim_{t\to0}\frac{\sin t\cos a + \sin a\cos t - \sin a}t$ and then she said that we apply limit only to $\sin a\cos t$ so it becomes $\sin a\cos0$ which is $\sin a$ and then $\lim_{t\to0}\frac{\sin t\cos a+\sin a-\sin a}t = \lim_{t\to0}\frac{\sin t\cos a}t$ as $\lim_{t\to0}\frac{\sin t}t$ gives $1$ so it is left $\cos a$.

I was wondering if you can apply limit just to a part as she did. I know you can separate but if you separate then you get $\lim_{t\to0}\frac{\sin t\cos a}t + \lim_{t\to0}\frac{\sin a\cos t}t - \lim_{t\to0}\frac{\sin a}t$ so its not the same as applying limit like that and separating because if you separate at $\lim_{t\to0}\frac{\sin a}t$ if you apply limit it becomes $\frac{\sin a}0$ so is it correct to do solve it as she did , I never seen it before so I am confused ?