To evaluate
$$\lim_{n\to\infty} \left[ \left( \int_{0}^{1} e^{nx(1-x)} \ dx \right)^\frac{1}{n} + \left( \int_{0}^{1} e^{\frac{x(1-x)}{n}} \ dx \right)^{n}\right]$$
I have used the M.V.T to give an upper bound $2e^\frac{1}{4}$ and a lower bound $2$ for the answer, but I am not able to get the correct answer. I feel that it is associated with L'Hopital's rule and the dominated convergence theorem, but these theorems may not lead to the right way.
Best Answer
For the first term:
Using $x(1-x) \le 1/4$ on $[0, 1]$, we have $$ \int_0^1 \mathrm{e}^{nx(1-x)}\,\mathrm{d} x \le \int_0^1 \mathrm{e}^{n/4}\,\mathrm{d} x = \mathrm{e}^{n/4}. $$
Using $x(1 - x) \ge 1/4 - 1/n^2$ on $[1/2 - 1/n, 1/2 + 1/n]$ ($n > 2$), we have $$\int_0^1 \mathrm{e}^{nx(1-x)}\,\mathrm{d} x \ge \int_{1/2-1/n}^{1/2 + 1/n} \mathrm{e}^{nx(1-x)}\,\mathrm{d} x \ge \int_{1/2-1/n}^{1/2 + 1/n} \mathrm{e}^{n(1/4 - 1/n^2)}\,\mathrm{d} x = \frac{2}{n} \mathrm{e}^{n(1/4 - 1/n^2)}.$$
Thus, we have $$\frac{2^{1/n}}{n^{1/n}}\mathrm{e}^{1/4 - 1/n^2} \le \left(\int_0^1 \mathrm{e}^{nx(1-x)}\,\mathrm{d} x\right)^{1/n} \le \mathrm{e}^{1/4}.$$
By the squeeze theorem, we have $$\lim_{n\to \infty} \left(\int_0^1 \mathrm{e}^{nx(1-x)}\,\mathrm{d} x\right)^{1/n} = \mathrm{e}^{1/4}.$$
For the second term:
Using $1 + u \le \mathrm{e}^u \le 1 + u + u^2$ on $[0, 1]$, we have $$1 + x(1-x)/n \le \mathrm{e}^{x(1-x)/n} \le 1 + x(1-x)/n + x^2(1-x)^2/n^2.$$
Thus, we have $$\left(1 + \frac{1}{6n} \right)^n\le \left(\int_0^1 \mathrm{e}^{x(1-x)/n}\,\mathrm{d} x \right)^n\le \left(1 + \frac{1}{6n} + \frac{1}{30n^2}\right)^n.$$
By the squeeze theorem, we have $$\lim_{n\to \infty} \left(\int_0^1 \mathrm{e}^{x(1-x)/n}\,\mathrm{d} x\right)^n = \mathrm{e}^{1/6}.$$