$f(x)=e^{-x},x>0$ Find $\lim_{n\rightarrow\infty}P(X_n<2+\log_en)$ Let $X_1,X_2....X_n$ be i.i.d random variables with pdf
$f(x)=e^{-x},x>0$
if $X_n=\max(X_1,X_1....X_n)$ then $\lim_{n\rightarrow\infty}P(X_n<2+\log_en)$
I have calculated Pdf of $X_n$ Which is $n(1-e^{-x})^{n-1}e^{-x}$
As $n$ goes large our distribution is Normal but I am not able to find the expectation. How do I solve this problem?
 A: I am assuming you meant $Y_n = \max(X_1,\dots,X_n)$, as $X_n = \max(X_1,\dots,X_n)$ does not really make sense.
You can compute the cdf exactly, without having to go through the pdf: note that for every $t$, $Y_n \leq t$ if, and only if, $X_i \leq t$ for every $1\leq i \leq n$ (by definition of the maximum). Then, using independence,
$$\begin{align}
\mathbb{P}\{ Y_n \leq 2+ \log n\} 
&= 
\mathbb{P}\{ \forall 1\leq i\leq n,\; X_i \leq 2+ \log n\}
= \prod_{i=1}^n \mathbb{P}\{  X_i \leq 2+ \log n\}\\
&= \mathbb{P}\{  X_1 \leq 2+ \log n\}^n
\end{align}$$
the last equality as all r.v.'s are identically distributed. But since $X_1$ has an exponential distribution, its CDF is $\mathbb{P}\{  X_1 \leq t\} = 1-e^{-t}$ for all $t\geq 0$, and so
$$
\mathbb{P}\{ Y_n \leq 2+ \log n\}  
= \left(1-e^{-(2+ \log n)}\right)^n
= \left(1-\frac{e^{-2}}{n}\right)^n
$$
Can you conclude by computing $\lim_{n\to\infty}\left(1-\frac{e^{-2}}{n}\right)^n$?
Remark: the above outline is quite general, and useful when dealing with the maximum of independent random variables.
A: $P(X_n<2+log \, \, n)=[P(X_1<2+log \, \,  n)]^{n}=(1-e^{-(2+log n)})^{n}=(1-\frac 1 {e^{2}n})^{n} \to e^{-e^{-2}}$ since $(1+\frac x n)^{n} \to e^{x}$ for any real number $x$.
