Conditions on a sequence of functions to satisfy a certain simple! condition in limit Let $f_n(x):[0,1]\to [0,1]$ be a sequence of continuous functions 
such that $f_n(x)\leq 1-x$ and $\int_0^1 f_n(x)\frac{1}{1-x} dx=\frac{1}{n}$. I am interested to know
what extra conditions I must impose on $f_n(.)$ to have the following property.
$$
\liminf_{n\to\infty} \frac{\int_{0}^1 f_n(x)dx}{\int_0^1 f_n(x)\frac{1}{1-x}dx} = \liminf_{n\to\infty} n \int_{0}^1 f_n(x)dx\geq \delta>0.
$$
for some $\delta>0$ (I need a guarantee that it does not converge to zero).
For example, one possible condition can be  $f_n(x)=0$ for $0<\alpha<x<1$, then
the above relation is always larger than $1-\alpha$. However, I want to know if
I assume that $f_n(.)$ are analytic functions, what conditions should they have?
Can I say something about the derivatives?
Let me give an example of analytic functions. Let $f_n(x)=\frac{1}{n(e-1)}(1-x)e^{(1-x)}$. Then, the above fraction is larger than 1/(e-1) for all $n$.
 A: For convenience, I changed variables to $g_n(x) = \frac{n \, f_n(x)}{1-x}$. Then, the functions $g_n$ satisfy
$$0 \le g_n \le n
\qquad\text{and}\qquad \int_0^1 g_n \, dx = 1.$$
Now, we show that your desired condition
$$\limsup \int_0^1 g_n(x) \, (1-x) \, dx \ge \delta$$
for some $\delta > 0$
is equivalent to
$$\exists \alpha, \beta \in (0,1) : \limsup \int_\alpha^1 g_n \le \beta.$$
Roughly speaking, this means, that not all mass clusters at $x = 1$.
I hope that this condition is simple enough for your purpose ;)
$\Leftarrow$:
From my condition and $\int g_n = 1$, we know
$$\liminf \int_0^\alpha g_n \, dx = 1- \limsup \int_\alpha^1 g_n \,dx \ge 1-\beta.$$
This yields $$\liminf \int_0^1 g_n \, (1-x) \, dx
\ge \liminf\int_0^\alpha g_n \, (1-x) \, dx \ge (1-\beta)\,(1-\alpha) > 0.$$
$\Rightarrow$:
Assume that my condition does not hold. This means,
for every $\alpha, \beta \in (0,1)$ and $N \in \mathbb N$, there exists $n \ge N$ with
$$\int_\alpha^1 g_n \ge \beta.$$
This shows
$$\int_0^1 g_n \, (1-x) \, dx
\le \int_0^\alpha g_n \, dx + \int_\alpha^1 g_n \, (1-\alpha) \, dx
\le (1-\beta) + (1-\alpha).$$
This, however, shows
$$\limsup \int_0^1 g_n \, dx = 0.$$
