Calculus - Finding limit (NOT L'Hopital's Rule): $\lim_{x \to 1^-}\frac{x^2+x+\sin({\pi\over 2}x)-3}{x-1}$ How do I find this limit?
$$\displaystyle{\lim_{x \to 1^-}}\frac{x^2+x+\sin({\pi \over 2}x)-3}{x-1}$$
I am unable to factor the numerator to get rid of the denominator. Can someone please help? Thank you!
Is there any other way to get the answer besides using L'Hopital's Rule?

Here we do not use Hopital or derivatives.
Note that 
$$x^2+x+\sin({\pi \over 2}x)-3=(x+2)(x-1)+\sin({\pi \over 2}-{\pi \over 2}(1-x))-1\\=(x+2)(x-1)+\cos({\pi \over 2}(1-x))-1\\
=(x+2)(x-1)-2\sin^2(\frac{\pi}{4}(1-x))$$
because $1-\cos(t)=2\sin^2(t/2)$.
Hence, as $x\to 1$,
$$\frac{x^2+x+\sin({\pi \over 2}x)}{(x-1)}=x+2-2\sin(\frac{\pi}{4}(1-x))\cdot \frac{\sin(\frac{\pi}{4}(1-x))}{(x-1)}\to 1+2+0=3$$
where we used the fact that 
$$\lim_{x\to 1}\frac{\sin(\frac{\pi}{4}(1-x))}{(x-1)}=-\frac{\pi}{4}\cdot \lim_{x\to 1}\frac{\sin(\frac{\pi}{4}(1-x))}{\frac{\pi}{4}(1-x)}=-\frac{\pi}{4}.$$
