<s>[INST] 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 ?
 [/INST] Just another way using Taylor series.
Built around $t=a$, Taylor series are $$\cot(t)=\cot (a)- \left(1+\cot ^2(a)\right)(t-a)+O\left((t-a)^2\right)\tag1$$ $$\sin(t)=\sin (a)+ \cos (a)(t-a)+O\left((t-a)^2\right)\tag2$$ So, replacing by the proper values $$\cot \left(\frac{\pi  x}{4}\right)=1-\frac{\pi}{2}   (x-1)+O\left((x-1)^2\right)$$ $$\sin(\pi x)=-\pi  (x-1)+O\left((x-1)^2\right)$$ from which it is easy to conclude.
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