A calculus problem by Rajyawardhan Singh

Since $\lim_{x \to 0+}x^x = 1$ , for any $\varepsilon > 0$ we can find $\delta > 0$ such that $|x^x - 1| < \varepsilon$ for $0 < x < \delta$ . Find $N \in \mathbb{N}$ with $N > \delta^{-1}$ . If $n \ge N$ then $\frac{1-\varepsilon}{k} = (1 - \varepsilon)n^k \int_0^{1/n} x^{k-1}\,dx \; < \; n^k \int_0^{1/n} x^{x+k-1}\, dx \; < \; (1 + \varepsilon)n^k\int_0^{1/n}x^{k-1}\,dx \; = \; \frac{1+\varepsilon}{k}$ for any $k > 0$ . Thus we deduce that $f(k) \; = \; \lim_{n \to \infty} n^k \int_0^{1/n} x^{x+k-1}\,dx \; = \; k^{-1}$ for all $k > 0$ , making the required answer $\boxed{5}$ .

Let z=1/t then t will tend to zero and then apply lopital rule and use lim z->0 z^z =1