<s>[INST] Trigonometric Limits - solution needed: $\lim_{x\to π/2} \frac{1-\sin x+\cos x}{\sin 2x -\cos x}$ how to solve this problem? (without using l'Hopital rule)
$$\lim_{x\to π/2} \frac{1-\sin x+\cos x}{\sin 2x -\cos x}$$
thanks for helping.
 [/INST] It is much simpler to put $x = \dfrac{\pi}{2} - t$ so that we have $\sin x = \cos t, \cos x = \sin t$ and $\sin 2x = \sin (\pi - 2t) = \sin 2t$. Also as $x \to \pi/2$ we have $t \to 0$. Clearly we then see that $$\begin{aligned}L &= \lim_{x \to \pi/2}\frac{1 - \sin x + \cos x}{\sin 2x - \cos x}\\
&= \lim_{t \to 0}\frac{1 - \cos t + \sin t}{\sin 2t - \sin t}\\
&= \lim_{t \to 0}\frac{1 - \cos t + \sin t}{\sin t(2\cos t - 1)}\\
&= \lim_{t \to 0}\frac{1 - \cos t + \sin t}{\sin t\cdot 1}\\
&= \lim_{t \to 0}\frac{1 - \cos t}{\sin t} + 1\\
&= \lim_{t \to 0}\frac{1 - \cos t}{t^{2}}\cdot t\cdot \frac{t}{\sin t} + 1\\
&= \frac{1}{2}\cdot 0\cdot 1 + 1 = 1\end{aligned}$$
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