TITLE: How should I solve this quesiton on Integration?
QUESTION [7 upvotes]: I want to answer this question:
If
$$
\lim_{n \to\infty} n^k \int_0^{1/n} x^{x+k-1} dx = f(k)
$$
for $k \in \mathbb N$, what is
$$
\left[\frac{1}{f(5)}\right],
$$
where square brackets denote the greatest integer function (i.e., ceil)?
I tried it by substituting $t = x + k - 1$, but got stuck.

REPLY [10 votes]: As innisfree suggested in his comment, this is an ideal case for the squeeze theorem.
Since $n \in \mathbb{N}$, we have $x \in [0,1/n] \subseteq [0,1]$ so the larger the exponents, the smaller the number. Thus
$$\int_0^{1/n}x^{1/n + k - 1} \,dx < \int_0^{1/n} x^{x+k-1} \,dx < \int_0^{1/n} x^{0+k-1} \,dx$$
$$\frac{1}{(1/n+k)n^k} < \int_0^{1/n} x^{x+k-1} \,dx < \frac{1}{kn^k}$$
$$\frac{1}{1/n+k} < n^k\int_0^{1/n} x^{x+k-1} \,dx < \frac{1}{k}$$
and taking the limit as $n \to \infty$ we find that 
$$\lim_{n \to \infty} n^k\int_0^{1/n} x^{x+k-1} \,dx = \frac{1}{k}$$