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For the personal research I would want to know the value of the limits of the following form:

$$\lim_{n\to \infty}\left( \int_0^1 (x^n+P(x))^ndx\right)^{\frac{1}{n}}$$

where $P(x)=x^m$ and m is a real number non-zero and positive.

What I managed to find was the fact that $$ (x^n+P(x))^n\to 0 $$ and I obtained the case $0^0$, which is unsolvable, however the geogebra and desmos confirmed to me that the limit is going to be $0$, but I have to prove it still.

I denoted $$L(m)=\lim_{n \to \infty}\left( \int_0^1 (x^n+x^m)^ndx\right)^{\frac{1}{n}}$$

I was thinking of applying Newton binomial Theorem but even though it was extremely labourious I could not be able to make any significant progress. I also tried because $x^\frac{1}{n}$ is concave, the Jensen inequility for integrals, which proved useless as well. What should I do?

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2 Answers 2

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The limit is equal $2.$ Indeed for $n\ge m$ we have $$ \left (\int\limits_0^1 (x^n+x^m)^n\,dx \right )^{1/n}\ge \left (\int\limits_0^1 (2x^n)^n\,dx \right )^{1/n}=2\left (\int\limits_0^1 x^{n^2}\,dx \right )^{1/n}={2\over (1+n^2)^{1/n}}\ge {2^{1-1/n}\over (n^{1/n})^2}.$$ On the other hand $$\left (\int\limits_0^1 (x^n+x^m)^n\,dx \right )^{1/n}\le \left (\int\limits_0^1 2^n\,dx \right )^{1/n}=2.$$

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  • $\begingroup$ I think there is an error with the signs of the first inequality, starts with $\geq$ and ends with $\leq$. $\endgroup$
    – Zaragosa
    Mar 10 at 22:09
  • $\begingroup$ @Zaragosa Thanks, I have corrected that. $\endgroup$ Mar 10 at 22:13
  • $\begingroup$ When you take limit in the first inequality it doesn't really come out 2, so I think it would be better to leave it at $\frac{2}{(1+n^2)^{1/n}}$, because $\displaystyle\lim_{n\to +\infty}\frac{2}{(1+n^2)^{1/n}}=\frac{2}{\displaystyle\lim_{n\to +\infty}e^{\frac{ln(1+n^2)}{n}}}$=2. $\endgroup$
    – Zaragosa
    Mar 10 at 22:20
  • $\begingroup$ $1+n^2 \leq 2n^2$ would do. $\endgroup$
    – Diger
    Mar 10 at 22:21
  • $\begingroup$ @Diger I had in mind exactly that inequality, but I have typed carelessly on my mobile phone while watching tennis. $\endgroup$ Mar 10 at 22:58
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I'll try to complete my answer later because I'm at work now, but I think this may give some better lights: Let be $F(n,m)=\displaystyle\left( \int_0^1 (x^n+x^m)^ndx\right)^{\frac{1}{n}}$. It is easy to see that $$ \int_0^1 (x^n+x^m)^ndx=\sum_{k=0}^n\frac{{n\choose k}}{n^2-nk+mk+1}\leq \sum_{k=0}^n{n\choose k}=2^n. $$ Hence it's easy to see that $F(1,m)<F(2,m)<\dots<F(n,m)<F(n+1,m)<\dots\leq2$. The idea to follow can be given from here. Although this is an upper bound, it does not have to be the smallest upper bound, this is what I will try when I finish working. Even if we get a subsequence $(F(n_k,m))_k$ convergent for $2$, then the whole sequence $(F(n,k))_n$ also converges for $2$, this is my idea to take advantage of the central term of a Newton binomial with even exponent. Another idea that I haven't been able to test yet is how $F(n,m)$ is monotonic and bounded so it is convergent for $L$, so we can take the natural logarithm and do L'Hospital, it will come out with the digamma function but it could be a more effective way believe.

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