Let $f$ be integrable on $[0,\infty$).
Show that lim$_{n \rightarrow \infty}$ $\int_0^{\infty}$ $f(x) \frac{x}{n+x}$ $dx$ = $0$.
Ok so I want to say I can take the limit inside, but this is a consequence of DCT or MCT, so to apply DCT, this is what I have come up with so far:
define $f_n$ = $f(x)$ $\cdot$ $\frac{x}{n+x}$ (my intuition from previous measure theory problems is that I need this step)
then {$f_n\}$ is a sequence of measurable functions,
but are the $f_n$ bounded above $f(x)$ ?
*Note by assumption we have $\int_0^{\infty}$ $f(x)$ $dx < \infty$.
Using all of this am I able to somehow deduce that
lim$_{n \rightarrow \infty}$ $\int_0^{\infty}$ $f(x) \frac{x}{n+x}$ $dx$ = $\int_0^{\infty}$ $f(x)$ lim$_{n \rightarrow \infty}$ $\frac{x}{n+x}$ $dx$ = $\int_0^{\infty}$ $f(x)$ $dx$ $\int_0^{\infty}$ lim$_{n \rightarrow \infty}$ $\frac{x}{n+x}$ $dx$
I am VERY hesitant on my above two equalities, specially the last one, where am I wrong or need input? Thanks in advance and sorry if I am missing something obvious.
I will do this from scratch. First, integrability reads $\int |f(x)|<\infty$. Now, set $f_n(x)= f(x)\frac{x}{n+x}$, and observe that, $|f_n(x)|\leq |f(x)|$ point wise, as $\frac{x}{n+x}\leq 1$ holds always, whenever $x\geq 0$, and $n\geq 1$.
Now, also note that, for any fixed $x$, $\lim_{n\to\infty}f_n(x)=0$, point wise. Hence, by DCT, $\lim_n \int f_n(x)\; dx=\int \lim_{n\to\infty}f_n(x)\; dx = 0$, as $\lim_n f_n(x)=0$.