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?
 A: One may recall that, for any differentiable function $f$ near $a$, one has
$$
\lim_{x \to a^-}\frac{f(x)-f(a)}{x-a}=f'(a^-)
$$ then observing that
$$
\frac{x^2+x+\sin({\pi\over 2}x)-3}{x-1}=\frac{\left(x^2+x+\sin({\pi\over 2}x)\right)-\left(1^2+1+\sin({\pi\over 2}\cdot1)\right)}{x-1}
$$ one gets that the sought limit is equal to
$$
f'(1^-)=\left.2x+1+\frac \pi 2 \cos \frac{\pi x}{2} \right|_{x \to 1^-}=3.
$$
A: Note that we can write
$$\begin{align}
\frac{x^2+x+\sin(\pi x/2)-3}{x-1}&=(x+2)+\frac{\sin(\pi x/2)-1}{x-1}\\\\
&=(x+2)-2\frac{\sin^2(\pi(x-1)/4)}{x-1}\\\\
&=(x+2)-\frac{\pi^2}{8}\left(\frac{\sin(\pi(x-1)/4)}{\pi (x-1)/4}\right)^2(x-1)
\end{align}$$
Inasmuch as $\lim_{x\to 1^-}\frac{\sin(\pi(x-1)/4)}{\pi (x-1)/4}=1$, the limit of interest is $3$.
A: 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 
$$\lim_{x\to 1}\frac{\sin(\frac{\pi}{4}(1-x))}{(x-1)}=-\frac{\pi}{4}\cdot \lim_{x\to 1}\frac{\sin(\frac{\pi}{4}(1-x))}{\frac{\pi}{4}(1-x)}=-\frac{\pi}{4}.$$
A: 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.
