Evaluate $\lim\limits_{n \to \infty} \int_{1-\epsilon}^1 \ln(1+x+...+x^{n-1})dx$ 
Evaluate $$\lim_{n \to \infty} \int_{1-\epsilon}^1 \ln(1+x+...+x^{n-1})dx$$
  for $0<\epsilon<1$.

The problem is at the point $x=1$, because there the formula $$1+x+...+x^{n-1}=\frac{1-x^n}{1-x}$$ can't be applied. I think of using the mean value theorem for integrals, so there would be $c_n \in [1-\epsilon,1]$ so that $$\int_{1-\epsilon}^1 \ln(1+x+...+x^{n-1})dx=\epsilon\ln(1+c_n+...+c_n^{n-1})$$
but taking limits gives an indeterminate form.
I do not know Lebesgue's dominated convergence theorem. I would appreciate an elementary proof.

You can replace the upper limit with $k$ and apply $\lim_{k\rightarrow 1}$ to the whole integral. Then that formula can be applied.
So the integral becomes
$$\int_{c}^{k} (\log(1-x^n) - \log (1-x))dx$$
Since $x<1$ in the given interval $\log(1-x^n)$ is $0$ as n tends to infinity.
So 
\begin{align*}
& =-\int_{c}^{k} \log (1-x))dx \\
& =(1-x)\log(1-x)-(1-x) \\
\end{align*}
from $c$ to $k$
$$=(1-k)\log(1-k)-(1-k)-((1-c)\log(1-c)-(1-c))$$
The limit of first term is $0$ as $k\rightarrow 1$. The second term is
$$=(1-c)(1-\log(1-c))$$
Now $c=1-\varepsilon$ like in the question
$$=\varepsilon(1-\log\varepsilon)$$
