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

Question

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 ?

Answer 1

HINT: use that $$\sin(x)-\sin(a)=-2 \sin \left(\frac{a}{2}-\frac{x}{2}\right) \cos \left(\frac{a}{2}+\frac{x}{2}\right)$$

Answer 2

You can consider $$ \sin x-\sin a=\sin((x-a)+a)=\sin(x-a)\cos a+\cos(x-a)\sin a $$ and you can rewrite your limit as $$ \lim_{x\to a}\left(\frac{\sin(x-a)}{x-a}\cos a-\frac{1-\cos(x-a)}{x-a}\sin a\right) $$ Now the standard limits $$ \lim_{t\to0}\frac{\sin t}{t}=1 \qquad \lim_{t\to0}\frac{1-\cos t}{t}=0 $$ solve the problem.

You indeed cannot disregard the second summand; however, it has limit $0$, so it doesn't contribute to the final limit.

On the other hand, you are computing the derivative of the sine at $a$, so “avoiding derivatives” seems like an oxymoron.