$X_1, X_2, \dots , X_n$ ~ i.i.d $U[0,1]$
a) Find the mean and variance of $lnX_1$.
b) Let $0\le a \lt b$. Find the $\lim_{n \to \infty} P(a \le (X_1,X_2, \dots ,X_n)^{(n^ -0.5)}*e^{ ( n^{0.5})}\le b)$ in terms of $a$ and $b$.
For part a since it is a uniform distribution I know that it will be equal to $1 \times (\frac{1}{(1-0)}) \times ln(1) = 0$. I tried to integrate this from $0$ to $1$ multiplying $x$ in order to find the expected value, but it is giving me the answer $0$. Same goes for variance.
For part $b$, I am stuck completely.
Help needed. Thank you.
Partial answer
mean of $\ln X_{1}$:
$\mathbb{E}\ln X_{1}=\lim_{\varepsilon\downarrow0}\int_{\varepsilon}^{1}\ln xdx=-1-\lim_{\varepsilon\downarrow0}\left(\varepsilon\ln\varepsilon-\varepsilon\right)=-1$.
Here $x\ln x-x$ is the antiderivative of $\ln x$ and $\lim_{\varepsilon\downarrow0}\varepsilon\ln\varepsilon=0$.
Note that $\ln X_{1}$ only takes negative values on interval $(0,1]$ and is not defined on $0$.
variance of $\ln X_{1}$:
$\mathbb{E}\ln^{2}X_{1}=\lim_{\varepsilon\downarrow0}\int_{\varepsilon}^{1}\ln^{2}xdx=2-\lim_{\varepsilon\downarrow0}\left(\varepsilon\ln^{2}\varepsilon-2\varepsilon\ln\varepsilon+2\varepsilon\right)=2$.
Here $x\ln^{2}x-2x\ln x+2x$ is the antiderivative of $\ln^{2}x$ and $\lim_{\varepsilon\downarrow0}\varepsilon\ln^{2}\varepsilon=0$.
This leads to $\text{Var}\left(\ln X_{1}\right)=\mathbb{E}\ln^{2}X_{1}-\left(\mathbb{E}\ln X_{1}\right)^{2}=2-(-1)^{2}=2-1=1$.