Evaluating $\lim\limits_{n \to \infty}\frac{\int_0^{1/2} x^{nx}\,\mathrm dx}{\int_{1/2}^1 x^{nx}\,\mathrm dx}$ Let $f_n(x):=x^{nx}=(x^x)^n$ and $f_n(0)=1.$ It's easy to see, for any fixed $n$, $f_n(x)$ is decreasing over $[0,1/e]$ and increasing over $[1/e,1]$.
Therefore for $0\le x\le 1/2$, it holds that$$e^{-n/e}=f_n(1/e)\le f_n(x)\le f_n(0)=1,$$and
for $1/2 \le x\le 1$, it holds that$$2^{-n/2}=f_n(1/2)\le f_n(x)\le f_n(1)=1.$$
Hence
$$\frac{1}{2}e^{-n/e}\le \int_0^{\frac{1}{2}} f_n(x)\,\mathrm dx\le \frac{1}{2},\qquad\frac{1}{2}\cdot2^{-n/2}\le \int_{\frac{1}{2}}^1 f_n(x)\,\mathrm dx\le \frac{1}{2}.$$
But this seems to be helpless.
 A: Let $x^n = y$, then both integrals become
$$I(a,b) \equiv \int_a^b x^{nx}\:dx = \frac{1}{n}\int_{a^n}^{b^n}y^{y^{\frac{1}{n}}-1+\frac{1}{n}}\:dy$$
and the limit is
$$L = \lim_{n\to\infty}\frac{I\left(0,\frac{1}{2}\right)}{I\left(\frac{1}{2},1\right)} = \lim_{n\to\infty}\frac{\int_{0}^{\frac{1}{2^n}}y^{y^{\frac{1}{n}}-1+\frac{1}{n}}\:dy}{\int_{\frac{1}{2^n}}^1 y^{y^{\frac{1}{n}}-1+\frac{1}{n}}\:dy} \to \frac{\int_0^0 dy}{\int_0^1 dy} = \frac{0}{1}=0$$
by dominated convergence.
A: $\lim_{n \to \infty} \frac{\int_{0}^{1/2} e^{nx \ln(x)} dx}{\int_{1/2}^{1} e^{nx\ln(x)} dx} $.
$x\ln(x)$ has its supremum at $x=0$ in $(0,1/2)$ and its supremum at $x=1$ in $(1/2,1)$, with both having that value $0$.
So by Laplace's method, this limit is equal to $\sqrt{\frac{\lim_{x \to 1}|d^2/dx^2(x \ln(x))|}{\lim_{x \to 0}|d^2/dx^2 (x \ln(x))|}}=0$.
A: Define $I_n = \displaystyle\int_{0}^{1/2}x^{nx}\,dx$ and $J_n = \displaystyle\int_{1/2}^{1}x^{nx}\,dx$. Trivially, $I_n \ge 0$ and $J_n \ge 0$ for all $n$.
Since $0 \le 1+\ln x \le 1$ for all $x \in [\tfrac{1}{2},1]$, we have $$J_n = \int_{1/2}^{1}x^{nx}\,dx \ge \int_{1/2}^{1}x^{nx}(1+\ln x)\,dx = \left[\dfrac{1}{n}x^{nx}\right]_{1/2}^{1} = \dfrac{1-2^{-n/2}}{n}.$$
Next, pick $\delta \in (0,\tfrac{1}{4})$, and split $I_n$ into two pieces: $$I_n = \int_{0}^{\delta}x^{nx}\,dx + \int_{\delta}^{1/2}x^{nx}\,dx.$$
Since $1+\ln \delta \le 1 + \ln x \le 0$ for all $x \in (0,\delta]$, we have $$\int_{0}^{\delta}x^{nx}\,dx \le \int_{0}^{\delta}\dfrac{x^{nx}(1+\ln x)}{1+\ln \delta}\,dx = \left[\dfrac{x^{nx}}{n(1+\ln \delta)}\right]_{x \to 0^{+}}^{x = \delta} = \dfrac{\delta^{n\delta}-1}{n(1+\ln \delta)} \le \dfrac{1}{n(\ln \tfrac{1}{\delta}-1)}$$
Since $x^x \le \max\{\delta^\delta,\tfrac{1}{\sqrt{2}}\} = \delta^\delta$ for all $x \in [\delta,\tfrac{1}{2}]$ (we implicitly used the fact that $0 < \delta < 1/4$), we have $$\displaystyle\int_{\delta}^{1/2}x^{nx}\,dx \le \left(\dfrac{1}{2}-\delta\right)\delta^{n\delta}.$$
Hence, $$I_n = \int_{0}^{\delta}x^{nx}\,dx + \int_{\delta}^{1/2}x^{nx}\,dx \le \dfrac{1}{n(\ln \tfrac{1}{\delta}-1)} + \left(\dfrac{1}{2}-\delta\right)\delta^{n\delta}$$ for any $\delta \in (0,\tfrac{1}{4})$ and any integer $n \ge 0$.
Then, we have $$0 \le \displaystyle\limsup_{n \to \infty}\dfrac{I_n}{J_n} \le \limsup_{n \to \infty}\dfrac{\tfrac{1}{n(\ln(1/\delta)-1)} + \left(\tfrac{1}{2}-\delta\right)\delta^{n\delta}}{\tfrac{1-2^{-n/2}}{n}} = \dfrac{1}{\ln \tfrac{1}{\delta}-1}$$ for any $\delta \in (0,\tfrac{1}{4})$. Now, take the limit as $\delta \to 0^+$ to get $0 \le \displaystyle\limsup_{n \to \infty}\dfrac{I_n}{J_n} \le 0$, i.e. $\displaystyle\limsup_{n \to \infty}\dfrac{I_n}{J_n} = 0$. Then, since $\dfrac{I_n}{J_n} \ge 0$ for all $n$, we have that $\displaystyle\lim_{n \to \infty}\dfrac{I_n}{J_n}$ exists and is $0$.
