## Problem 4.

a) Compute

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

b) Let $k \geq 1$ be an integer. Compute

$$
\lim _{n \rightarrow \infty} n^{k+1} \int_{0}^{1}\left(\frac{1-x}{1+x}\right)^{n} x^{k} d x
$$
Solution. $a$ ) The limit equals $\frac{1}{2}$. The result follows immediately from $b$ ) for $k=0$.

b) The limit equals $\frac{k!}{2^{k+1}}$. We have, by the substitution $\frac{1-x}{1+x}=y$, that

$$
\begin{aligned}
n^{k+1} \int_{0}^{1}\left(\frac{1-x}{1+x}\right)^{n} x^{k} d x & =2 n^{k+1} \int_{0}^{1} y^{n}(1-y)^{k} \frac{d y}{(1+y)^{k+2}} \\
& =2 n^{k+1} \int_{0}^{1} y^{n} f(y) d y
\end{aligned}
$$

where

$$
f(y)=\frac{(1-y)^{k}}{(1+y)^{k+2}}
$$

We observe that

$$
f(1)=f^{\prime}(1)=\cdots=f^{(k-1)}(1)=0
$$

We integrate $k$ times by parts $\int_{0}^{1} y^{n} f(y) d y$, and by (7) we get

$$
\int_{0}^{1} y^{n} f(y) d y=\frac{(-1)^{k}}{(n+1)(n+2) \ldots(n+k)} \int_{0}^{1} y^{n+k} f^{(k)}(y) d y
$$

One more integration implies that

$$
\begin{aligned}
\int_{0}^{1} y^{n} f(y) d y= & \frac{(-1)^{k}}{(n+1)(n+2) \ldots(n+k)(n+k+1)} \\
& \times\left(\left.f^{(k)}(y) y^{n+k+1}\right|_{0} ^{1}-\int_{0}^{1} y^{n+k+1} f^{(k+1)}(y) d y\right) \\
= & \frac{(-1)^{k} f^{(k)}(1)}{(n+1)(n+2) \ldots(n+k+1)} \\
& +\frac{(-1)^{k+1}}{(n+1)(n+2) \ldots(n+k+1)} \int_{0}^{1} y^{n+k+1} f^{(k+1)}(y) d y
\end{aligned}
$$

It follows that

$$
\lim _{n \rightarrow \infty} 2 n^{k+1} \int_{0}^{1} y^{n} f(y) d y=2(-1)^{k} f^{(k)}(1)
$$

since

$$
\lim _{n \rightarrow \infty} \int_{0}^{1} y^{n+k+1} f^{(k+1)}(y) d y=0
$$

$f^{(k+1)}$ being continuous and hence bounded. Using Leibniz's formula we get that

$$
f^{(k)}(1)=(-1)^{k} \frac{k!}{2^{k+2}}
$$

and the problem is solved.