Finding Limit Using Taylor Polynomial 
Find the limit $$\displaystyle{\lim_{x\to4}}\frac{\left(1-\cos \left( x-4 \right)\right)^4\,\ln \left( x-3 \right)}{\left(e^{\left(x-4\right)^2}-1\right)^2\,\sin ^5\left(\pi \,x\right)}$$
  Using Taylor Polynomial and Peano reminder

So first thing would be to move $\lim_{x\to 4}$ to $\lim_{x\to 0}$
So we get $$\displaystyle{\lim_{x\to0}}\frac{\left(1-\cos \left( x \right)\right)^4\,\ln \left( x+1 \right)}{\left(e^{\left(x\right)^2}-1\right)^2\,\sin ^5\left(\pi \,x\right)}$$
using Talyor expansion we get:
$$\displaystyle{\lim_{x\to0}}\frac{1-(1-\frac{x^2}{2}+\frac{x^4}{4}+o(x^6))^4\cdot x-\frac{x^2}{2}-\frac{x^3}{3}+o(x^6)}{x^2+\frac{x^4}{2}+o(x^6)\cdot (\pi^5 x^5-\frac{5\pi^7 x^7}{6}+\frac{23\pi^9x^9}{72}+o(x^{11})}$$
how should I processed from here? and why does the taylor expansion of $\sin ^5\left(\pi \,x\right)= $ this?
 A: Using Taylor for this simple limit is not a good idea. However if one does wish to use the Taylor series it can be done by a little simplification at first (similar to what I present below) and then using first term (i.e. upto $x^{1}$) of the Taylor series for $\sin x, \log(1 + x), e^{x}$ and for $\cos x$ you need to go upto $x^{2}$.
On the other hand without Taylor, we can put $x = 4 + h$ so that $h \to 0$ and then we have
\begin{align}
L & = \lim_{x \to 4}\frac{(1 - \cos(x - 4))^{4}\log (x - 3)}{\left(e^{(x - 4)^{2}} - 1\right)^2\sin ^{5}(\pi x)}\notag\\
&= \lim_{h \to 0}\frac{(1 - \cos h)^{4}\log (1 + h)}{\left(e^{h^{2}} - 1\right)^2\sin ^{5}(\pi(4 + h))}\notag\\
&= \lim_{h \to 0}\frac{(1 - \cos h)^{4}\log (1 + h)}{\left(e^{h^{2}} - 1\right)^2\sin ^{5}(\pi h)}\text{ (may use Taylor after this step)}\tag{1}\\
&= \lim_{h \to 0}\frac{(1 - \cos h)^{4}}{h^{8}}\cdot\frac{\log (1 + h)}{h}\cdot\frac{h^{9}}{\left(e^{h^{2}} - 1\right)^2\sin ^{5}(\pi h)}\notag\\
&= \frac{1}{2^{4}}\lim_{h \to 0}\frac{h^{9}}{\left(e^{h^{2}} - 1\right)^2\sin ^{5}(\pi h)}\notag\\
&= \frac{1}{2^{4}}\lim_{h \to 0}\frac{h^{4}}{\left(e^{h^{2}} - 1\right)^2}\cdot\frac{(\pi h)^{5}}{\sin ^{5}(\pi h)}\frac{1}{\pi^{5}}\notag\\
&= \frac{1}{2^{4}}\cdot\frac{1}{\pi^{5}} = \frac{1}{16\pi^{5}}
\end{align}
Here we have used standard limits
$$\lim_{x \to 0}\frac{\sin x }{x} = 1, \lim_{x \to 0}\frac{1 - \cos x}{x^{2}} = \frac{1}{2}, \lim_{x \to 0}\frac{\log(1 + x)}{x} = 1, \lim_{x \to 0}\frac{e^{x} - 1}{x} = 1$$
If we wish to use Taylor after step $(1)$ above then we can note that
\begin{align}
\cos h &= 1 - \frac{h^{2}}{2} + o(h^{2})\notag\\
e^{h^{2}} &= 1 + h^{2} + o(h^{2})\notag\\
\sin(\pi h) &= \pi h + o(h)\notag\\
\log(1 + h) &= h + o(h)\notag
\end{align}
It follows from the above Taylor expansions that
\begin{align}
\lim_{h \to 0}\frac{1 - \cos h}{h^{2}} &= \frac{1}{2}\notag\\
\lim_{h \to 0}\frac{\log(1 + h)}{h} &= 1\notag\\
\lim_{h \to 0}\frac{h^{2}}{e^{h^{2}} - 1} &= 1\notag\\
\lim_{h \to 0}\frac{\pi h}{\sin(\pi h)} &= 1\notag
\end{align}
Thus continuing from equation $(1)$ we have
\begin{align}
L &= \lim_{h \to 0}\frac{(1 - \cos h)^{4}\log (1 + h)}{\left(e^{h^{2}} - 1\right)^2\sin ^{5}(\pi h)}\notag\\
&= \lim_{h \to 0}\left(\frac{(1 - \cos h)}{h^{2}}\right)^{4}\cdot\frac{\log (1 + h)}{h}\left(\frac{h^{2}}{e^{h^{2}} - 1}\right)^{2}\left(\frac{\pi h}{\sin(\pi h)}\right)^{5}\frac{1}{\pi^{5}}\notag\\
&= \frac{1}{2^{4}}\cdot 1\cdot 1^{2}\cdot 1^{5}\cdot\frac{1}{\pi^{5}} = \frac{1}{16\pi^{5}}
\end{align}
