Finding a limit without using L'Hospital's rule I encountered an issue for finding the limit for a function, one small step I just can't get. 
I know that this is classic case for L'Hospital's however our calculus course didn't reach it and I can not use it for this one.
$
2.3 .m,n\in\mathbb{N}\quad\underset{x\rightarrow1}{\lim}\frac{x^{m}-1}{x^{n}-1}$
$x\neq1$
\begin{aligned}\underset{x\rightarrow1}{\lim}\frac{x^{m}-1}{x^{n}-1}=\underset{x\rightarrow1}{\lim}\frac{x^{m}-1^{m}}{x^{n}-1^{n}}=\underset{x\rightarrow1}{\lim}\frac{\left(x-1\right)\sum_{i=0}^{m-1}x^{i}\cdot1^{m-i-1}}{\left(x-1\right)\sum_{i=0}^{m-1}x^{j}\cdot1^{n-j-1}}=\\
\underset{x\rightarrow1}{\lim}\frac{\left(m-1\right)x^{m-1}}{\left(n-1\right)x^{n-1}}=\underset{x\rightarrow1}{\lim}\frac{mx^{m-1}-x^{m-1}}{nx^{n-1}-x^{n-1}}
\end{aligned}
And here I stuck... 
know I need to reach $\frac{m}{n}$ but how can I proceed...?!!? nothing to cancel out here... 

Define $\displaystyle L(m,n) = \underset{x\rightarrow1}{\lim}\frac{x^{m}-1}{x^{n}-1}.$ You wish to find the value of $L(m,n)$ for $m,n\in\mathbb N$.
It’s easy to see that $L(k,k)=1$, and if $L(m,n)$ exists and is nonzero, then $L(m,n)=\dfrac{1}{L(n,m)}$. Now notice that
$$L(m+1,n)=\underset{x\rightarrow1}{\lim}\frac{x^{m+1}-1}{x^{n}-1}=\underset{x\rightarrow1}{\lim}\frac{x\left(x^{m}-1\right)+(x-1)}{x^{n}-1}=L(m,n)+L(1,n),$$
so $L[m,n]$ satisfies the recurrence relation
$L[m,n]={\Big\{}
\begin{array}{cc}
  \ & 
    \begin{array}{cc}
      1 & m=n \\
      \dfrac{1}{L[n,m]} & m<n \\
      L(m-1,n)+L(1,n) & m>n
    \end{array}
\end{array}
$
This recurrence relation has a unique solution $L[m,n]=\dfrac{m}{n}$.
