Integral limit struggle: lim[n->infty]n-th root of (int[0,1](1+x^n)^n dx)
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We are given to evaluate:
\(\displaystyle \displaystyle \lim_{n\to\infty}\left(\sqrt[n]{\int_0^1 \left(1+x^n\right)^n\,dx}\right)\)
Let \(\displaystyle f(x)=1+x^n\) and observe that for all \(\displaystyle x\in\left[0,1\right]\) and \(\displaystyle n\in\mathbb{N}\), we have \(\displaystyle 0<f(x)\).
Now, let's define:
\(\displaystyle M\equiv\sup_{x \in [0,1]} f(x)\)
Observe then that we must have:
\(\displaystyle \displaystyle \left(\int_0^1 f(x)^n\,dx\right)^{\frac{1}{n}} \le \left(\int_0^1 M^n\,dx\right)^{\frac{1}{n}} = M\)
Thus, we conclude:
\(\displaystyle \displaystyle \limsup_{n\to\infty}\left(\left(\int_0^1 f(x)^n\,dx\right)^{\frac{1}{n}}\right)\le M\)
Can you now find a way to show:
\(\displaystyle M\le\displaystyle \liminf_{n\to\infty}\left(\left(\int_0^1 f(x)^n\,dx\right)^{\frac{1}{n}}\right)\) ?
\(\displaystyle \displaystyle \lim_{n\to\infty}\left(\sqrt[n]{\int_0^1 \left(1+x^n\right)^n\,dx}\right)\)
Let \(\displaystyle f(x)=1+x^n\) and observe that for all \(\displaystyle x\in\left[0,1\right]\) and \(\displaystyle n\in\mathbb{N}\), we have \(\displaystyle 0<f(x)\).
Now, let's define:
\(\displaystyle M\equiv\sup_{x \in [0,1]} f(x)\)
Observe then that we must have:
\(\displaystyle \displaystyle \left(\int_0^1 f(x)^n\,dx\right)^{\frac{1}{n}} \le \left(\int_0^1 M^n\,dx\right)^{\frac{1}{n}} = M\)
Thus, we conclude:
\(\displaystyle \displaystyle \limsup_{n\to\infty}\left(\left(\int_0^1 f(x)^n\,dx\right)^{\frac{1}{n}}\right)\le M\)
Can you now find a way to show:
\(\displaystyle M\le\displaystyle \liminf_{n\to\infty}\left(\left(\int_0^1 f(x)^n\,dx\right)^{\frac{1}{n}}\right)\) ?
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Just to follow up, here's the solution I had in mind:
We are given to evaluate:
\(\displaystyle \displaystyle \lim_{n\to\infty}\left(\sqrt[n]{\int_0^1 \left(1+x^n\right)^n\,dx}\right)\)
Let \(\displaystyle \displaystyle f(x)=1+x^n\) and observe that for all \(\displaystyle \displaystyle x\in\left[0,1\right]\) and \(\displaystyle \displaystyle n\in\mathbb{N}\), we have \(\displaystyle \displaystyle 0<f(x)\).
Now, let's define:
\(\displaystyle \displaystyle M\equiv\sup_{x \in [0,1]} f(x)\)
Observe then that we must have:
\(\displaystyle \displaystyle \left(\int_0^1 f(x)^n\,dx\right)^{\frac{1}{n}} \le \left(\int_0^1 M^n\,dx\right)^{\frac{1}{n}}= M\)
Thus, we conclude:
\(\displaystyle \displaystyle \limsup_{n\to\infty}\left(\left(\int_0^1 f(x)^n\,dx\right)^{\frac{1}{n}}\right)\le M\)
Now let \(\displaystyle \alpha\) be any non-negative real number strictly less than \(\displaystyle M\). By definition, there must be some \(\displaystyle x_0\in[0,1]\) such that \(\displaystyle f\left(x_0\right)=M\). By continuity of \(\displaystyle f\), we can find an interval \(\displaystyle (c,d) \subset [0,1]\) such that \(\displaystyle f(x)>\alpha\) for all \(\displaystyle x\in(c,d)\). Then, for every \(\displaystyle n\), we have:
\(\displaystyle \displaystyle \left(\int_0^1 f(x)^n\,dx\right)^{\frac{1}{n}}\ge\left(\int_c^d f(x)^n\,dx\right)^{\frac{1}{n}}\ge\left(\int_c^d \alpha^n\,dx\right)^{\frac{1}{n}}=\alpha(d-c)^{\frac{1}{n}}\)
Taking limits, there results:
\(\displaystyle \displaystyle \liminf_{n\to\infty}\left(\left(\int_0^1 f(x)^n\,dx\right)^{\frac{1}{n}}\right)\ge\lim_{n\to\infty}\left(\alpha(d-c)^{\frac{1}{n}}\right)=\alpha\)
Given that \(\displaystyle \alpha\) is an arbitrary real number strictly less than \(\displaystyle M\), the above implies:
\(\displaystyle \displaystyle \liminf_{n\to\infty}\left(\left(\int_0^1 f(x)^n\,dx\right)^{\frac{1}{n}}\right)\ge M\)
Thus, we have:
\(\displaystyle \displaystyle M\le\liminf_{n\to\infty}\left(\left(\int_0^1 f(x)^n\,dx\right)^{\frac{1}{n}}\right)\le\limsup_{n\to\infty}\left(\left(\int_0^1 f(x)^n\,dx\right)^{\frac{1}{n}}\right)\le M\)
And this implies:
\(\displaystyle \displaystyle \lim_{n\to\infty}\left(\left(\int_0^1 f(x)^n\,dx\right)^{\frac{1}{n}}\right)=M\)
For \(\displaystyle f(x)=1+x^n\), we find \(\displaystyle M=2\).
We are given to evaluate:
\(\displaystyle \displaystyle \lim_{n\to\infty}\left(\sqrt[n]{\int_0^1 \left(1+x^n\right)^n\,dx}\right)\)
Let \(\displaystyle \displaystyle f(x)=1+x^n\) and observe that for all \(\displaystyle \displaystyle x\in\left[0,1\right]\) and \(\displaystyle \displaystyle n\in\mathbb{N}\), we have \(\displaystyle \displaystyle 0<f(x)\).
Now, let's define:
\(\displaystyle \displaystyle M\equiv\sup_{x \in [0,1]} f(x)\)
Observe then that we must have:
\(\displaystyle \displaystyle \left(\int_0^1 f(x)^n\,dx\right)^{\frac{1}{n}} \le \left(\int_0^1 M^n\,dx\right)^{\frac{1}{n}}= M\)
Thus, we conclude:
\(\displaystyle \displaystyle \limsup_{n\to\infty}\left(\left(\int_0^1 f(x)^n\,dx\right)^{\frac{1}{n}}\right)\le M\)
Now let \(\displaystyle \alpha\) be any non-negative real number strictly less than \(\displaystyle M\). By definition, there must be some \(\displaystyle x_0\in[0,1]\) such that \(\displaystyle f\left(x_0\right)=M\). By continuity of \(\displaystyle f\), we can find an interval \(\displaystyle (c,d) \subset [0,1]\) such that \(\displaystyle f(x)>\alpha\) for all \(\displaystyle x\in(c,d)\). Then, for every \(\displaystyle n\), we have:
\(\displaystyle \displaystyle \left(\int_0^1 f(x)^n\,dx\right)^{\frac{1}{n}}\ge\left(\int_c^d f(x)^n\,dx\right)^{\frac{1}{n}}\ge\left(\int_c^d \alpha^n\,dx\right)^{\frac{1}{n}}=\alpha(d-c)^{\frac{1}{n}}\)
Taking limits, there results:
\(\displaystyle \displaystyle \liminf_{n\to\infty}\left(\left(\int_0^1 f(x)^n\,dx\right)^{\frac{1}{n}}\right)\ge\lim_{n\to\infty}\left(\alpha(d-c)^{\frac{1}{n}}\right)=\alpha\)
Given that \(\displaystyle \alpha\) is an arbitrary real number strictly less than \(\displaystyle M\), the above implies:
\(\displaystyle \displaystyle \liminf_{n\to\infty}\left(\left(\int_0^1 f(x)^n\,dx\right)^{\frac{1}{n}}\right)\ge M\)
Thus, we have:
\(\displaystyle \displaystyle M\le\liminf_{n\to\infty}\left(\left(\int_0^1 f(x)^n\,dx\right)^{\frac{1}{n}}\right)\le\limsup_{n\to\infty}\left(\left(\int_0^1 f(x)^n\,dx\right)^{\frac{1}{n}}\right)\le M\)
And this implies:
\(\displaystyle \displaystyle \lim_{n\to\infty}\left(\left(\int_0^1 f(x)^n\,dx\right)^{\frac{1}{n}}\right)=M\)
For \(\displaystyle f(x)=1+x^n\), we find \(\displaystyle M=2\).