<s>[INST] How do I evaluate $ \lim_{x \to \pi/2}\frac {\sin x-1}{(1+\cos 2x)}$without using L'Hospital's rule? How do I evaluate this: $$\lim_{x \to \pi/2}\displaystyle \frac {\sin x-1}{(1+\cos 2x)}$$? without using L'Hospital's rule.
Attempt : I used trigonomeitric transformation and I  used Taylor series I got this :
$$\lim_{x \to \pi/2}\displaystyle \frac {\sin x-1}{(1+\cos 2x)} =
\lim_{y\to 0}\displaystyle \frac {\cos y-cos²y}{2\cos² (2y+\pi)}-\frac{1}{2}$$
where $y=x-\frac{\pi}{2}$, but by this way i got :
$$\lim_{x \to \pi/2}\displaystyle \frac {\sin x-1}{(1+\cos 2x)}=\frac{-1}{2}$$ and by L'Hospital's rule is :$\frac{-1}{4}$
Then where is my problem if I'm true
Thank you for any help .
 [/INST] $$\lim_{x\to \frac{\pi}{2}} \frac {\sin x-1}{(1+\cos 2x)} = \lim_{x\to \frac{\pi}{2}} \frac {\sin x-1}{(1+\cos^2x-\sin^2x)} = \\
= \lim_{x\to \frac{\pi}{2}} \frac {\sin x-1}{(1+(1-\sin^2x)-\sin^2x)} = \\
= \lim_{x\to \frac{\pi}{2}} \frac {\sin x-1}{2(1-\sin^2x)} = \\
= \lim_{x\to \frac{\pi}{2}} \frac {\sin x-1}{2(1-\sin x)(1+\sin x)} = \\
= \lim_{x\to \frac{\pi}{2}} \frac {-1}{2(1+\sin x)} = -\frac{1}{4}\\$$
Using Taylor for $x_0 = \frac{\pi}{2}$...
$$\sin(x) = 1 - (\pi/2 - x)^2/2 + \ldots$$
$$\cos(2x) = -1 + 2(\pi/2 - x)^2 + \ldots$$
Then:
$$\lim_{x\to \frac{\pi}{2}} \frac {\sin x-1}{(1+\cos 2x)} = \lim_{x\to \frac{\pi}{2}} \frac { 1 - (\pi/2 - x)^2/2-1+ \ldots}{(1-1 + 2(\pi/2 - x)^2 + \ldots)} = \\
=\lim_{x\to \frac{\pi}{2}} \frac {- (\pi/2 - x)^2/2+ \ldots}{2(\pi/2 - x)^2 + \ldots} = -\frac{1}{4}$$
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