$\ \lim_{x\to0} \frac{x}{\cos(\frac{\pi}{2}-x)} $ without L'Hospital's rule 
How do I find the following?$$\ \lim_{x\to0} \frac{x}{\cos(\frac{\pi}{2}-x)} $$

I have tried to use trig identities :
$$ \frac{x}{\cos(\frac{\pi}{2}-x)} = \frac{x}{\cos\frac{\pi}{2}\cos x+\sin\frac{\pi}{2}\sin x} =  $$
Can't really see anything out of that?
edit- so based on Math lover hint :
$$\ \frac{x}{\cos \frac{\pi}{2}\cos x + \sin \frac{\pi}{2}\sin x} = \frac{x}{0 \cdot \cos x + 1\cdot \sin x } = \frac{x}{\sin x} = \frac{0}{0} = 0 $$

As an alternative by $f(x)=\cos\left(\frac{\pi}{2}-x\right)$ from the definition of derivative
$$\lim_{x\to0} \frac{x}{\cos\left(\frac{\pi}{2}-x\right)}=\lim_{x\to0} \frac{x-0}{\cos\left(\frac{\pi}{2}-x\right)-\cos\left(\frac{\pi}{2}\right)}=\frac1{f'(0)}=1$$
