How to calculate the limit: $\lim _{ x\to1 } \frac { nx^{ n+1 }-(n+1)x^{ n }+1 }{ (e^{ x }-e)\sin(\pi x) } $ How to calculate the limit?
$$\lim_{ x\to 1 } \frac { nx^{ n+1 }-(n+1)x^{ n }+1 }{ (e^{ x }-e)\sin(\pi x) } $$ 
Is binomial theorem needed ?
 A: $$\lim_{x\to1} \frac{nx^{n+1}-(n+1)x^n+1}{(e^x -e)\sin(\pi x)}$$
Using l'hopitals since $\frac{0}{0}$ is undefined:
$$\lim_{x\to1} \frac{n(n+1)x^{n}-n(n+1)x^{n-1}}{e^x\sin(\pi x)+\pi(e^x-e)\cos(\pi x)}$$
$$\lim_{x\to1} \frac{n(n+1)(x^{n}-x^{n-1})}{e^x\sin(\pi x)+\pi(e^x-e)\cos(\pi x)}$$
L'hopitals again
$$\lim_{x\to1} \frac{n(n+1)(nx^{n-1}-(n-1)x^{n-2})}{e^x\sin(\pi x)+\pi e^x\cos(\pi x)-\pi^2 (e^x-e)\sin(\pi x)+\pi e^x \cos(\pi x)} = \frac{n(n+1)}{-2\pi e^x} $$
Therefore:The answer is 
$$\lim_{x\to1} \frac{nx^{n+1}-(n+1)x^n+1}{(e^x -e)\sin(\pi x)}= -\frac{n(n+1)}{2\pi e} $$
A: Rewrite the function as
$$\left(x-1\over e^x-e\right)\left(x-1\over\sin\pi x \right)\left(nx^{n+1}-(n+1)nx^n+1\over(x-1)^2 \right)$$
Now run L'Hopital on the three pieces individually.  (You'll need to L'Hopitate the third piece twice.)
A: Suppose $n$ is a positive integer first. In that case we can put $x = 1 + h$ and let $h \to 0$. The limit evaluation is as follows
\begin{align}
L &= \lim_{h \to 0}\frac{n(1 + h)^{n + 1} - (n + 1)(1 + h)^{n} + 1}{(e^{1 + h} - e)\sin\pi(1 + h)}\notag\\
&= -\frac{1}{e}\lim_{h \to 0}\frac{n(1 + h)^{n + 1} - (n + 1)(1 + h)^{n} + 1}{(e^{h} - 1)\sin\pi h}\notag\\
&= -\frac{1}{e}\lim_{h \to 0}\frac{n(1 + h)^{n + 1} - (n + 1)(1 + h)^{n} + 1}{h^{2}}\cdot\frac{h}{e^{h} - 1}\cdot\frac{\pi h}{\sin\pi h}\cdot\frac{1}{\pi}\notag\\
&= -\frac{1}{\pi e}\lim_{h \to 0}\frac{n(1 + h)^{n + 1} - (n + 1)(1 + h)^{n} + 1}{h^{2}}\tag{1}\\
&= -\frac{1}{\pi e}\lim_{h \to 0}\dfrac{n\left(1 + (n + 1)h + \dfrac{n(n + 1)}{2}h^{2} + \cdots\right) - (n + 1)\left(1 + nh + \dfrac{n(n - 1)}{2}h^{2} + \cdots\right) + 1}{h^{2}}\notag\\
&= -\frac{1}{\pi e}\lim_{h \to 0}\dfrac{\dfrac{n^{2}(n + 1)}{2}h^{2} - \dfrac{n(n^{2} - 1)}{2}h^{2} + o(h^{2})}{h^{2}}\notag\\
&= -\frac{1}{\pi e}\cdot\frac{n^{2} + n}{2} = -\frac{n(n + 1)}{2\pi e}\notag
\end{align} Note that the ellipsis ($\ldots$) used in above derivation represents a finite (because $n$ is positive integer) number of terms in powers of $h$ starting with $h^{3}$ and hence they are combined and written as $o(h^{2})$.
If $n$ is not a positive integer then we have to start from equation $(1)$ and apply L'Hospital's Rule. Thus
\begin{align}
L &= -\frac{1}{\pi e}\lim_{h \to 0}\frac{n(1 + h)^{n + 1} - (n + 1)(1 + h)^{n} + 1}{h^{2}}\notag\\
&= -\frac{1}{\pi e}\lim_{h \to 0}\frac{n(n + 1)(1 + h)^{n} - n(n + 1)(1 + h)^{n - 1}}{2h}\notag\\
&= -\frac{n(n + 1)}{2\pi e}\lim_{h \to 0}\frac{h(1 + h)^{n - 1}}{h}\notag\\
&= -\frac{n(n + 1)}{2\pi e}\notag
\end{align}
A: Hint: Use L'Hôpital's rule:
\begin{eqnarray}
\lim_{ x\to 1 } \frac { nx^{ n+1 }-(n+1)x^{ n }+1 }{ (e^{ x }-e)\sin(\pi x) } 
&=& \lim_{ x\to 1 } \frac { n(n+1)x^n-n(n+1)x^{n-1} }{ e^{ x }\sin(\pi x) + \pi(e^{ x }-e)\cos(\pi x) } \\
&=& \lim_{ x\to 1 } \frac { n^2(n+1)x^{n-1}-n(n+1)(n-1)x^{n-2} }{ e^{ x }\sin(\pi x) + 2\pi e^{ x }\cos(\pi x) - \pi^2(e^{ x }-e)\sin(\pi x) }
\end{eqnarray}
