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 ?
 A: \begin{align*}
lim_{ u \to 0}\frac{\frac{\tan( \frac{π}{4}(u+1))-1)}{\tan( \frac{\pi}{4}(u+1))}}{-\sin \pi u} &= lim_{ u \to 0}\frac{\frac{1+\tan\frac{\pi}{4} u}{1-\tan\frac{\pi}{4} u}-1}{-\sin \pi u} \\
&= lim_{ u \to 0}\frac{2\tan\frac{\pi}{4} u}{-\sin \pi u}\\
&=  lim_{ u \to 0}\frac{\frac{\frac{\pi}{2}\tan\frac{\pi}{4} u}{\frac{\pi}{4}u}}{-\frac{\sin \pi u}{u}}\\
&= -\frac{1}{2}
\end{align*}
A: $$\lim_{ x \to 1}\frac{\tan(\frac{\pi}{4}x)-1}{\tan(\frac{\pi}{4}x)\sin πx}=$$
$$\lim_{ x \to 1}\frac{\tan(\frac{\pi}{4}x)-\tan(\frac{\pi}{4})}{\tan(\frac{\pi}{4}x)\sin πx}=$$
$$\lim_{ x \to 1}\frac{\sin(\frac{\pi}{4}(x-1))}{\cos(\frac{\pi}{4}x)\cos(\frac{\pi}{4})\tan(\frac{\pi}{4}x)\sin(\pi-\pi x)}=$$
$$\lim_{ x \to 1}\frac{\sin(\frac{\pi}{4}(x-1))}{\cos(\frac{\pi}{4})\sin(\frac{\pi}{4}x)\sin\pi(1- x)}=$$
$$\lim_{ x \to 1}\frac{(\frac{\pi}{4}(x-1))}{\cos(\frac{\pi}{4})\sin(\frac{\pi}{4}x)\pi(1- x)}=-\frac12$$
A: 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 $\cos(2x) = \cos^2(x)-\sin^2(x)$
A: Let $\dfrac\pi4(x-1)=y\iff\dfrac{\pi x}4=y+\dfrac\pi4$
$$\lim_{x\to1}\dfrac{1-\cot\dfrac{\pi x}4}{\sin\pi x}=\lim_{y\to0}\dfrac{1-\cot\left(y+\dfrac\pi4\right)}{\sin(4y+\pi)}$$
$$=\lim_{y\to0}\dfrac{1-\dfrac{\cot y-1}{\cot y+1}}{-\sin4y}$$
$$=-\lim_{y\to0}\dfrac2{(\cot y+1)\sin4y}$$
$$=-\lim_{y\to0}\dfrac{2\sin y}{(\cos y+\sin y)\sin4y}$$
Can you take it from here?
A: With your attempt 
$$
1-\cot( \frac{π}{4}x)=1-\frac{1}{\tan( \frac{π}{4}x)}=\frac{\tan( \frac{π}{4}x)-1}{\tan( \frac{π}{4}x)}
$$
you can go further 
\begin{align}
\frac{1-\cot( \frac{π}{4}x)}{\sin\pi x}&=\frac{\tan( \frac{π}{4}x)-1}{\tan( \frac{π}{4}x)\sin\pi x}\\
&=\frac{\cos\frac{\pi}{4}x(\tan( \frac{π}{4}x)-1)}{\sin( \frac{π}{4}x)\sin\pi x}\\
&=\frac{\sin\frac{\pi}{4}x-\cos\frac{\pi}{4}x}{\sin( \frac{π}{4}x)\sin\pi x}\tag{1}
\end{align}
use product to sum formula for the denominator
$$\sin( \frac{π}{4}x)\sin\pi x=\frac{1}{2}\left[\cos\left(\frac{3}{4}\pi x\right)-\cos\left(\frac{5}{4}\pi x\right)\right]\tag{2}$$ 
by substituting $(2)$ in eq. $(1)$ compute the limit.
A: 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.
A: Let $f(x) = \cot (\pi x/4),g(x) = \sin (\pi x).$ The expression equals
$$\frac{f(1) - f(x)}{g(x)-g(1)}= -\frac{(f(x) - f(1))/(x-1)}{(g(x)-g(1))/(x-1)}.$$
By definition of the derivative, as $x\to 1,$ the numerator on the right $\to -f'(1),$ the denominator $\to g'(1).$ So the limit is $-f'(1)/g'(1),$ a simple computation. (And we are not using L'Hopital, thank you very much.)
