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?

Another possible way to do it.
Start changing variable $x=y+1$; this gives
$$\frac{x^2+x+\sin \left(\frac{\pi  x}{2}\right)-3}{x-1}=\frac{y^2+3 y+\cos \left(\frac{\pi  y}{2}\right)-1}{y}$$ Now, use Taylor expansion $$\cos(t)=1-\frac{t^2}{2}+O\left(t^4\right)$$ which gives $$\cos \left(\frac{\pi  y}{2}\right)=1-\frac{\pi ^2 y^2}{8}+O\left(y^4\right)$$ So $$\frac{y^2+3 y+\cos \left(\frac{\pi  y}{2}\right)-1}{y}=\frac{y^2 +3y-\frac{\pi ^2 y^2}{8}+O\left(y^4\right)}{y}=(1-\frac{\pi ^2 }{8})y+3+O\left(y^3\right)$$ which shows the limit and how it is approached.
