Question : $Let\space$ {$ \space { f_n : [0,1] \to R }\space $}${_{n=1}^{\infty}}$ be a sequence of non- negative continuous functions such that $f_n(1)=1$.
Statement 1 : Then, it is possible that $\lim_{n \to \infty}({ \int_{0}^{1}{f_n(x)\,dx}) }=0$
Statement 2 : Then, it is possible that $\lim_{n \to \infty}({ \int_{0}^{1}{exp((f_n(x))\,dx)} }=0$
Which of the following is the true answer ?
1.) both statements
2.) statement 1
3.) statement 2
4.) none of the statements
approach, For statement 1 :
Suppose $f_n(x) = x^n$
Case 1 : when x = 1
$f_n(1)=(1)^n= 1 $.
Thus, $f_1(x)=f_2(x)=f_3(x).... = 1$
$\lim_{n \to \infty}f_n(x)=1$ as $n\to \infty$
Case 2 : for all $ x \in [0,1)$
$|x|< 1$,
$\lim_{n \to \infty} x^n = 0,$ so $f_n(x) = x^n \to 0$ as $n \to \infty$
Conclusion : $\lim_{n \to \infty}({ \int_{0}^{1}{f_n(x)\,dx}) }= \lim_{n \to \infty}({ \int_{0}^{1}{x^n\,dx}) }$ = $\lim_{n \to \infty} [\frac{x^{n+1}}{n+1}]_{0}^{1}$= $\lim_{n \to \infty} \frac{1}{n+1}=0$
approach for statement 2 :
As $\space e^{f_n(x)} \ge 1$ ; it can be written that $\space\space \lim_{n \to \infty}({ \int_{0}^{1}{exp((f_n(x))\,dx)} }\ge\space\space \lim_{n \to \infty}({ \int_{0}^{1}{1\,dx}) } = 1$. Cant not equal zero .
My final answer : statement 1 is the only correct .
I wanted to confirm whether the approach is correct or not? And the answer?. Please clarify