Find the limit without L'hopital rule $\lim_{ x \to 1}\frac{1-\cot(\frac{π}{4}x)}{\sin πx}=$? Find the limit without L'hopital rule 
$$\lim_{ x \to 1}\frac{1-\cot(\frac{π}{4}x)}{\sin πx}=?$$
My Try:
$$1-\cot( \frac{π}{4}x)=1-\frac{1}{\tan( \frac{π}{4}x)}=\frac{\tan( \frac{π}{4}x)-1}{\tan( \frac{π}{4}x)}\\\sin (πx)=\sin (\pi-\pi x)=-\sin \pi(x-1)\\\lim_{ x \to 1}\frac{1-\cot(\frac{π}{4}x)}{\sin πx}=\lim_{ x \to 1}\frac{\frac{\tan( \frac{π}{4}x)-1}{\tan( \frac{π}{4}x)}}{-\sin \pi(x-1)}\\u=x-1⇒x=u+1\\\lim_{ u \to 0}\frac{\frac{\tan( \frac{π}{4}(u+1))-1)}{\tan( \frac{π}{4}(u+1))}}{-\sin \pi u}$$
now ?

An alternate method - 
Convert $1 - \cot{(\dfrac \pi4x)}$ to $\dfrac {\sin {(\dfrac \pi4x)}-\cos({\dfrac \pi4x})}{\sin (\dfrac \pi4x)}$.
Multiply divide by ${\sin {(\dfrac \pi4x)}+\cos({\dfrac \pi4x})}$ and simplify using the formulae 
$\sin (2x) = 2\cos(x)\sin(x)$ and $\cos(2x) = \cos^2(x)-\sin^2(x)$
