Problem 2. Let $k>1$ be a real number. Calculate:

(a) $L=\lim _{n \rightarrow \infty} \int_{0}^{1}\left(\frac{k}{\sqrt[n]{x}+k-1}\right)^{n} \mathrm{~d} x$.

(b) $\lim _{n \rightarrow \infty} n\left[L-\int_{0}^{1}\left(\frac{k}{\sqrt[n]{x}+k-1}\right)^{n} \mathrm{~d} x\right]$.
Proof. (a) The limit equals $\frac{k}{k-1}$.

Using the substitution $x=y^{n}$ we have that

$$
I_{n}=\int_{0}^{1}\left(\frac{k}{\sqrt[n]{x}+k-1}\right)^{n} \mathrm{~d} x=n k^{n} \int_{0}^{1}\left(\frac{y}{y+k-1}\right)^{n-1} \frac{\mathrm{d} y}{y+k-1}
$$

Using the substitution $\frac{y}{y+k-1}=t \Rightarrow y=\frac{(k-1) t}{1-t}$ we get, after some calculations, that

$$
I_{n}=n k^{n} \int_{0}^{\frac{1}{k}} \frac{t^{n-1}}{1-t} \mathrm{~d} t
$$

We integrate by parts and we have that

$$
I_{n}=\frac{k}{k-1}-k^{n} \int_{0}^{\frac{1}{k}} \frac{t^{n}}{(1-t)^{2}} \mathrm{~d} t
$$

It follows that $\lim _{n \rightarrow \infty} I_{n}=\frac{k}{k-1}$ since

$$
0<k^{n} \int_{0}^{\frac{1}{k}} \frac{t^{n}}{(1-t)^{2}} \mathrm{~d} t<\frac{k^{n+2}}{(k-1)^{2}} \int_{0}^{\frac{1}{k}} t^{n} \mathrm{~d} t=\frac{k}{(n+1)(k-1)^{2}}
$$

(b) The limit equals $\frac{k}{(k-1)^{2}}$.

We have that

$$
\frac{k}{k-1}-I_{n}=k^{n} \int_{0}^{\frac{1}{k}} \frac{t^{n}}{(1-t)^{2}} \mathrm{~d} t
$$

We integrate by parts and we have that

$$
\frac{k}{k-1}-I_{n}=\frac{1}{n+1} \cdot \frac{k}{(k-1)^{2}}-\frac{2 k^{n}}{n+1} \int_{0}^{\frac{1}{k}} \frac{t^{n+1}}{(1-t)^{3}} \mathrm{~d} t
$$

This implies that

$$
\begin{aligned}
\lim _{n \rightarrow \infty} n\left[\frac{k}{k-1}-\int_{0}^{1}\left(\frac{k}{\sqrt[n]{x}+k-1}\right)^{n} \mathrm{~d} x\right] & = \\
= & \lim _{n \rightarrow \infty}\left[\frac{n}{n+1} \cdot \frac{k}{(k-1)^{2}}-\frac{2 k^{n} n}{n+1} \int_{0}^{\frac{1}{k}} \frac{t^{n+1}}{(1-t)^{3}} \mathrm{~d} t\right]
\end{aligned}
$$

Thus

$$
\lim _{n \rightarrow \infty} n\left[\frac{k}{k-1}-\int_{0}^{1}\left(\frac{k}{\sqrt[n]{x}+k-1}\right)^{n} \mathrm{~d} x\right]=\frac{k}{(k-1)^{2}}
$$

since

$$
0<k^{n} \int_{0}^{\frac{1}{k}} \frac{t^{n+1}}{(1-t)^{3}} \mathrm{~d} t<\frac{k^{n+3}}{(k-1)^{3}} \int_{0}^{\frac{1}{k}} t^{n+1} \mathrm{~d} t=\frac{k}{(k-1)^{3}(n+2)}
$$