find $\lim _{x\to \infty }\left(\frac{n^{k-1}}{n^k-\left(n-1\right)^k}\right)=\frac{1}{2005}$ the value of k $\lim _{x\to \infty }\left(\frac{n^{k-1}}{n^k-\left(n-1\right)^k}\right)=\frac{1}{2005}$
What is the value of k in the given expression?
When I expand the term (n-1) to the k-th power I get n^k-n^k.. 

$$\lim _{x\to \infty }\left(\frac{n^{k-1}}{n^k-\left(n-1\right)^k}\right)=\lim _{x\to \infty }\frac{1}{n}\left(\frac{1}{1-\left(1-\frac{1}{n}\right)^k}\right)=\lim _{x\to \infty }\left(\frac{\frac{1}{n}}{1-\left(1-\frac{1}{n}\right)^k}\right)$$
the limit is $\frac{0}{0}$, so you can use the L'hopital Rule
$$\lim _{x\to \infty }\left(\frac{-\frac{1}{n^2}}{-k\left(1-\frac{1}{n}\right)^{k-1}(\frac{1}{n^2})}\right)=\frac{1}{k}$$
so the $k$
$$k={2005}$$
