$$\lim\limits_{x\to \frac{\pi}{2}}\frac{\sin(x)-1}{x-\frac{\pi}{2}}$$
I tried to compare this function with the derivate definition formula $$\lim\limits_{x\to a} \frac{f(x)-f(a)}{x-a}$$
And I did find the correct solution, but I'm not sure that this isn't the same as using L'Hôpital... Please if somebody can help me I would be really grateful! Thanks for the help!
$$\lim_{x\to \pi/2}\frac{\sin(x)-1}{x-\pi/2}=\lim_{z\to 0}\frac{\sin(z+\pi/2)-1}{z}=\lim_{z\to 0}\frac{\cos(z)-1}{z}=\lim_{z\to 0}\frac{-2\sin^2\frac{z}{2}}{z}=\color{red}{0}$$ since $\lim_{z\to 0}\frac{\sin\frac{z}{2}}{\frac{z}{2}}=1$.
Here's another approach that uses the definition of the derivative $$\lim\limits_{x\to \frac{\pi}{2}}\frac{\sin(x)-1}{x-\frac{\pi}{2}}$$ Let $t=x-\frac{\pi}{2}$, then $$\lim\limits_{t\to 0}\frac{\sin\left(t+\frac{\pi}{2}\right)-1}{t}$$ $$=\lim\limits_{t\to 0}\frac{\sin\left(\frac{\pi}{2}+t\right)-\sin\left(\frac{\pi}{2}\right)}{t}$$ $$=\frac{\mathrm d}{\mathrm dx}\sin(x)\Bigg|_{x=\frac{\pi}{2}}=\cos(x)\Bigg|_{x=\frac{\pi}{2}}$$ $$=\cos\left(\frac{\pi}{2}\right)=0$$ Note that this approach does not use L'hôpital's rule.
$$\frac{\sin x - 1}{x-\pi/2} = \frac{\sin x - \sin \pi/2}{x-\pi/2} \to \sin'(\pi/2) = \cos (\pi/2) = 0.$$
