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.

• Hint: $a^n-b^n=(a-b)(a^{n-1}+ba^{n-2}+...+b^{n-2}a+b^{n-1})$. May 23 '15 at 15:21

If you're getting $c/0$ when using L'Hospital's Rule, then the limit doesn't exist.
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 $0/0$.)
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.
If you use the binomial expansion or Taylor series, the numerator is something like $$\frac{1}{2} \left(m^2+m-2 n\right)(x-1)^2+\cdots$$