Limit without De L'Hospital: $\lim_{x\to \pi/2}\frac{\sin x-1}{2x-\pi}$ We must to calculate $$\lim_{x\to \pi/2}\frac{\sin x-1}{2x-\pi}.$$
It is very simple with l'Hospital's rule that the limit is zero. But what would happen if I didn't use l'Hospital's rule?
In fact if $t=x-\frac{\pi}2$ we have $x=t+\frac{\pi}2$ hence
$$\lim_{x\to \pi/2} \frac{\sin x-1}{2x-\pi}=\lim_{t\to 0}\frac{\sin \left(t+\frac{\pi}2\right)-\sin\frac{\pi}2}{2t}=\lim_{t\to 0}\frac{2\sin\left(t+\pi\right)\cos t}{2t}=$$$$-\lim_{t\to 0}\frac{\sin t\cos t}{t}=1$$
that is not $0$. What am I doing wrong? Any trivial mistakes?
 A: 
We must to calculate $$\lim_{x\to \pi/2}\frac{\sin x-1}{2x-\pi}$$

If we use a change of variables $\ h = x - \frac{\pi}{2},\ $then this becomes:
$$\lim_{h\to \ 0}\frac{\sin\left(\frac{\pi}{2} + h\right) - \sin\left(\frac{\pi}{2}\right)}{2h}$$
$$\overset{(*)}{=}\frac{1}{2}\ f'\left(\frac{\pi}{2}\right),$$
where $\ f(x) = \sin(x).\ $ All we have done at $\ (*)\ $is use the definition of $\ f'(x).$
But $\ f'\left(\frac{\pi}{2}\right) = 0,\ $ and therefore the answer is $\ \frac{1}{2} \times 0 = 0.$
$$$$
Also, from the comments, you went wrong here:

$$\lim_{t\to 0}\frac{\sin
\left(t+\frac{\pi}2\right)-\sin\frac{\pi}2}{2t}=\lim_{t\to
0}\frac{2\sin\left(t+\pi\right)\cos t}{2t}$$

This is incorrect. The correct formula is:
$\ \sin\alpha - \sin \beta = 2 \cos \frac{\alpha+\beta}{2} \sin\frac{\alpha - \beta}{2},\ $ which would get you:
$\ \sin\left(t + \frac{\pi}{2}\right) - \sin\left(\frac{\pi}{2}\right) = 2\cos \left( \frac{1}{2} \left( t+ \pi \right)\right)\sin\left( \frac{1}{2} t\right)$
A: Another way to rewrite it is
$$ \frac{\sin(t + \pi/2) - 1}{2t} = \frac{\sin t\cos(\pi/2) + \sin(\pi/2)\cos t - 1}{2t} = \frac{\cos t - 1}{2t}. $$
