\begin{align*}
2b &= \lim_n n \int_0^1 \frac{nx^{n-1}} {x+1} \mathrm dx - \frac n 2 \\
&= \lim_n n \int_0^1 \frac {\mathrm d (x^n)} {x+1} - \frac n2\\
&= \lim_n n \left. \frac {x^n} {1+x}\right|_0^1 + n \int_0^1 \frac {x^n \mathrm dx} {(1+x)^2} - \frac n 2\\
&= \lim_n \frac n {n+1} \cdot \left.\frac {x^{n+1}}{(x+1)^2}\right|_0^1 + \frac {2n}{n+1}\int_0^1 \frac {x^{n+1}}{(x+1)^3} \mathrm d x\\
&= \frac 14 + 2\lim_n \int_0^1 \frac {x^{n+1} \mathrm dx} {(x+1)^3}\\
&= \frac 14,
\end{align*}
where
$$
0 \gets \frac 18 \int_0^1 x^{n+1} \mathrm dx \leqslant \int_0^1 \frac {x^{n+1} \mathrm dx} {(1+x)^3} \leqslant \int_0^1 x^{n+1} \mathrm d x \to 0.
$$
UPDATE
The limit you gave at the very first is actually $b$, not $2b$.
As we have:
$$\sin^2(t)=\frac12(1-\cos(2t))$$
It is clear that the function being integrated has a periodicity of $\pi$. Hence every integral through a whole $\pi$ period will have the same value. So, we have that:
$$\int\limits_0^\pi\sin^2(t)dt=\int\limits_0^\pi\frac12(1-\cos(2t))dt=\frac\pi2$$
Then, if we break down the integration as the function is suggesting:
$$\int\limits_0^n\sin^2(t)dt=\int\limits_0^{\pi \lfloor\frac{n}{\pi}\rfloor}\sin^2(t)dt + \int\limits_{\pi \lfloor\frac{n}{\pi}\rfloor}^n\sin^2(t)dt=\frac\pi2 \lfloor\frac{n}\pi\rfloor + \int\limits_{\pi \lfloor\frac{n}{\pi}\rfloor}^n\sin^2(t)dt$$
Notice that the integrand is always positive, so we can easily find an lower and upper bound by excluding the left term or letting it complete another cycle, which means that:
$$\frac\pi2 \lfloor\frac{n}\pi\rfloor<\frac\pi2 \lfloor\frac{n}\pi\rfloor + \int\limits_{\pi \lfloor\frac{n}{\pi}\rfloor}^n\sin^2(t)dt<\frac\pi2 \lfloor\frac{n}\pi\rfloor+\frac\pi2$$
$$\frac1n \frac\pi2 \lfloor\frac{n}\pi\rfloor<U_n<\frac1n\left(\frac\pi2 \lfloor\frac{n}\pi\rfloor+\frac\pi2 \right)$$
So, if we let $n \to \infty$, as both sides of the inequality tend to $1/2$, we have that
$$\lim_{n\to\infty}U_n=\frac12$$
Best Answer
On the one hand, it's at least $\lim_{n\to\infty}\frac{1}{n}\int_0^1\ln(e^{nx})dx=\int_0^1xdx=\frac12$. On the other hand, it's at most $\lim_{n\to\infty}\frac{1}{n}\int_0^1\ln(2e^{nx})dx=\frac12+\lim_{n\to\infty}\frac{\ln 2}{n}\int_0^1dx=\frac12$.