# PROBLEM 1 

$$
\text { Calculate } \lim _{n \rightarrow \infty} n\left(\frac{1}{2}-\int_{0}^{1} \frac{x}{1+e^{-n x}} d x\right) \text {. }
$$

(Mathematical Gazette)
## Solution:

$$
\frac{1}{2}-\int_{0}^{1} \frac{x}{1+e^{-n x}} d x=\int_{0}^{1} x d x-\int_{0}^{1} \frac{x}{1+e^{-n x}} d x=\int_{0}^{1} \frac{x d x}{1+e^{n x}} \text{. We have .......................... 2p }
$$

![](https://cdn.mathpix.com/cropped/2024_06_07_db10151b2bb7cf77684cg-1.jpg?height=122&width=1260&top_left_y=1115&top_left_x=449)

It follows that $\quad 0 \leq \int_{0}^{1} \frac{x}{1+e^{n x}} \leq \int_{0}^{1} x e^{-n x} d x=\int_{0}^{1} x\left(-\frac{e^{-n x}}{n}\right)^{1} d x=-\left.x \frac{e^{-n x}}{n}\right|_{0} ^{1}+\int_{0}^{1} \frac{e^{-n x}}{n} d x$

$$
=-\left.x \frac{e^{-n x}}{n}\right|_{0} ^{1}+\int_{0}^{1} \frac{e^{-n x}}{n} d x=-\frac{e^{-n}}{n}-\left.\frac{e^{-n x}}{n^{2}}\right|_{0} ^{1}=-\frac{e^{-n}}{n}-\frac{e^{-n}}{n^{2}}+\frac{1}{n^{2}} . \quad 2 \text{p}
$$

Then $\quad 0 \leq \lim _{n \rightarrow \infty} n\left(\frac{1}{2}-\int_{0}^{1} \frac{x}{1+e^{-n x}} d x\right) \leq \lim _{n \rightarrow \infty}\left(-\frac{1}{e^{n}}-\frac{1}{e^{n} n}+\frac{1}{n}\right)=0$, thus

$$
\lim _{n \rightarrow \infty} n\left(\frac{1}{2}-\int_{0}^{1} \frac{x}{1+e^{-n x}} d x\right)=0
$$

$1 \text{p}$