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.
 A: take $x=\frac\pi2-t$, since $\cos x=1-2\sin^2\frac x2, \sin 2x=2\sin x\cos x$, we have 
$$\lim_{x\to π/2} \frac{1-\sin x+\cos x}{\sin 2x -\cos x}=\lim_{x\to 0} \frac{1-\cos x+\sin x}{\sin 2x -\sin x}= \lim_{x\to 0} \frac{2\sin^2\frac x2+\sin x}{2\sin x\cos x -\sin x} $$
notice $\sin x=2\sin \frac x2\cos \frac x2$, so
$$\frac{2\sin^2\frac x2+\sin x}{2\sin x\cos x -\sin x} =\frac{2\sin^2\frac x2+2\sin \frac x2\cos \frac x2}{4\sin \frac x2\cos \frac x2\cos x -2\sin \frac x2\cos \frac x2}\\=\frac{2\sin\frac x2+2\cos \frac x2}{4\cos \frac x2\cos x -2\cos \frac x2} \to \frac2{4-2}=1(x\to 0)$$
A: $$\lim_{x\to\dfrac\pi2} \frac{1-\sin x+\cos x}{\sin 2x -\cos x}$$
$$=\lim_{x\to\dfrac\pi2} \frac{1-\sin x+\cos x}{\cos x}\cdot\frac1{\lim_{x\to\dfrac\pi2}(2\sin x-1)}$$
The second limit converges to $1$
$$\text{For the first , }\lim_{x\to\dfrac\pi2} \frac{1-\sin x+\cos x}{\cos x}=1+\lim_{x\to\dfrac\pi2}\dfrac{1-\sin x}{\cos x}$$
as $\displaystyle\cos x\ne0 \text{ as }x\ne\dfrac\pi2\text{ as }x\to\dfrac\pi2$
Observe that the main idea remained : to separate out all the terms from the limit which converges to non-zero finite numbers
Method $\#1:$
$$L=\lim_{x\to\dfrac\pi2}\frac{1-\sin x}{\cos x}=\lim_{x\to\dfrac\pi2}\frac{(1-\sin x)(1+\sin x)}{\cos x}\cdot\frac1{\lim_{x\to\dfrac\pi2}(1+\sin x)}$$
$$=\lim_{x\to\dfrac\pi2}\cos x\cdot\frac1{1+\sin\dfrac\pi2}=0\cdot\frac1{1+1}$$
Method $\#2:$
Using Double angle formula
$$L=\lim_{x\to\dfrac\pi2}\frac{1-\sin x}{\cos x}=\lim_{x\to\dfrac\pi2}\frac{\left(\cos\dfrac x2-\sin\dfrac x2\right)^2}{\cos^2\dfrac x2-\sin^2\dfrac x2}$$
$$=\lim_{x\to\dfrac\pi2}\frac{\cos\dfrac x2-\sin\dfrac x2}{\cos\dfrac x2+\sin\dfrac x2}$$
as $\displaystyle \cos\dfrac x2-\sin\dfrac x2\ne0\iff\tan\dfrac x2\ne1\iff\dfrac x2\frac\pi4\iff x\ne\dfrac\pi2$ as $x\to\dfrac\pi2$
$\displaystyle\implies L=\lim_{x\to\dfrac\pi2}\frac{\cos\dfrac x2-\sin\dfrac x2}{\cos\dfrac x2+\sin\dfrac x2}=0$
Method $\#3:$ Like Mark Bennet's answer use Weierstrass substitution 
Method $\#4:$ Like ziang chen's answer set $\displaystyle x=\dfrac\pi2-t$
A: 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 $$\begin{aligned}L &= \lim_{x \to \pi/2}\frac{1 - \sin x + \cos x}{\sin 2x - \cos x}\\
&= \lim_{t \to 0}\frac{1 - \cos t + \sin t}{\sin 2t - \sin t}\\
&= \lim_{t \to 0}\frac{1 - \cos t + \sin t}{\sin t(2\cos t - 1)}\\
&= \lim_{t \to 0}\frac{1 - \cos t + \sin t}{\sin t\cdot 1}\\
&= \lim_{t \to 0}\frac{1 - \cos t}{\sin t} + 1\\
&= \lim_{t \to 0}\frac{1 - \cos t}{t^{2}}\cdot t\cdot \frac{t}{\sin t} + 1\\
&= \frac{1}{2}\cdot 0\cdot 1 + 1 = 1\end{aligned}$$
A: Hint: Here is one way if nothing easier comes to mind - try the Weierstrass substitution $t=\tan \frac x2$ where $\sin x=\cfrac {2t}{1+t^2}, \cos x=\cfrac {1-t^2}{1+t^2}$ together with $\sin 2x = 2\sin x \cos x$
If you clear fractions you will get rational function of $t$ and you should be able to factorise numerator and denominator and cancel the problematic term. Since $\tan \frac {\pi}4 =1$ you should anticipate a common factor $t-1$ - since numerator and denominator are polynomials which come to zero at $t=1$.
A: here are some hints.

*sin 2x = 2 sin x cos x

*see if you can express the given fraction as two fractions instead of one.

*can you build a difference of squares?

