How to evaluate $\lim\limits_{x\to1}\frac{x^{m+1}-x^{n+1}+x^n-mx+m-1}{(x-1)^2}$? 
How to evaluate $$\lim\limits_{x\to1}\frac{x^{m+1}-x^{n+1}+x^n-mx+m-1}{(x-1)^2}$$

I used substitution $t=x-1 \Rightarrow t\rightarrow 0$
After this, the limit is:
$$\lim\limits_{t\to0}\frac{(t+1)^{m+1}-(t+1)^{n+1}+(t+1)^n-m(t+1)+m-1}{t^2},m,n \in \mathbb{N}$$
Then I used L'Hospital's rule, but I get $\frac{const.}{0}$
Can L'Hospital's rule be used, or some other method?
Thanks for replies.
 A: If you're getting $c/0$ when using L'Hospital's Rule, then the limit doesn't exist.
A: Hint: Use L'Hospital's rule twice, and you will get 2 in the denominator and be able to evaluate the limit. (After the first application I believe you will still get $0/0$.)
A: The numerator can be written
$$
x^{m+1}-1-x^{n+1}+x^n-mx+m=
(x-1)\biggl(\biggl(\sum_{k=0}^m x^k\biggr)-x^n-m\biggr)
$$
Set $f_m(x)=\sum_{k=0}^m x^k$, so your limit can be written as
$$
\lim_{x\to1}\frac{f_m(x)-x^n-m}{x-1}
$$
Note that $f_m(1)=m+1$, so we can apply l'Hôpital, giving
$$
\lim_{x\to1}(f_m'(x)-nx^{n-1})
$$
Since
$$
f_m'(x)=\sum_{k=1}^m kx^{k-1}
$$
we have
$$
f_m'(1)=1+2+\dots+m=\dots
$$
and it should be clear how to finish.
A: I think there is no truck here, it suffices to apply l'Hospital's rule
twice only
\begin{eqnarray*}
\lim_{x\rightarrow 1}\frac{x^{m+1}-x^{n+1}+x^{n}-mx+m-1}{(x-1)^{2}}
&=&\lim_{x\rightarrow 1}\frac{(m+1)x^{m}-(n+1)x^{n}+nx^{n-1}-m}{2(x-1)}\
(by\ HRule:\frac{0}{0}) \\
&=&\lim_{x\rightarrow 1}\frac{m(m+1)x^{m-1}-n(n+1)x^{n-1}+n(n-1)x^{n-2}}{2}\
\ (by\ HRule:\frac{0}{0}) \\
&=&\frac{m(m+1)-n(n+1)+n(n-1)}{2} \\
&=&\frac{m(m+1)-n^{2}-n+n^{2}-n}{2} \\
&=&\frac{m(m+1)}{2}-n.
\end{eqnarray*}
A: If you use the binomial expansion or Taylor series, the numerator is something like $$\frac{1}{2}  \left(m^2+m-2 n\right)(x-1)^2+\cdots$$
