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.. 

Using Taylor expansions, assuming $k \geq 1$ (so that things do tend to $\infty$ when $n\to\infty$):
$$\begin{align}
n^k - (n-1)^k &= n^k\left(1-\left(1-\frac{1}{n}\right)^k\right) = n^k\left(1-\left(1-\frac{k}{n} + o\left(\frac{1}{n}\right)\right)\right) \\&= kn^{k-1} + o\left(n^{k-1}\right)
\end{align}$$
so
$$
\frac{n^{k-1}}{n^k - (n-1)^k} = \frac{n^{k-1}}{kn^{k-1} + o(n^{k-1})} = \frac{1}{k+o(1)} \xrightarrow[n\to\infty]{}\frac{1}{k}
$$
and you can conclude by unicity of the limit.
