What is the limit of x->1 sinpi*x/(1-x)^2?
I want to solve it without using L'Hospital rule.
We are given to evaluate:
$\displaystyle \lim_{x\to1}\frac{\sin(\pi x)}{(1-x)^2}$
The first thing I am looking for is a way to apply the well-known result:
(1) $\displaystyle \lim_{u\to0}\frac{\sin(u)}{u}=1$
Seeing the denominator has $\displaystyle 1-x$ as a factor, this takes me to $\displaystyle \sin(\pi x)=\sin(\pi-\pi x)=sin(\pi(1-x))$. So we now have:
$\displaystyle \lim_{x\to1}\frac{\sin(\pi(1-x))}{(1-x)^2}$
Now, let's use the substitution $\displaystyle x=1-x$ and we have:
$\displaystyle \lim_{u\to0}\frac{\sin(\pi u)}{u^2}$
Now let's finish rewriting the expression so we can apply (1):
$\displaystyle \lim_{u\to0}\frac{\pi\sin(\pi u)}{\pi u^2}=\lim_{u\to0}\frac{\sin(\pi u)}{\pi u}\cdot\frac{\pi}{u}=\lim_{u\to0}\frac{\sin(\pi u)}{\pi u}\cdot\lim_{u\to0}\frac{\pi}{u}$
Applying (1), we are left with:
$\displaystyle \pi\lim_{u\to0}\frac{1}{u}$
Now, do we have:
$\displaystyle \lim_{u\to0^{-}}\frac{1}{u}=\lim_{u\to0^{+}}\frac{1}{u}$ ?