Let $$ I_{n} = \int_{0}^{1}\frac{x^{n}}{x^{n}+1}dx $$ I would like to know whether it's possible to prove that $$ \lim_{n \to \infty} I_{n} = 0 $$ using the mean theorem for integrals.
It is quite straightforward to prove the limit is 0 as follows from below:
$$ 0\leq I_{n} = \int_{0}^{1}\frac{x^{n}}{x^{n}+1}dx \leq \int_{0}^{1}x^{n}dx=\frac{1}{n+1}$$
Thus by the squeeze theorem we can conclude $$ \lim_{n \to \infty} I_{n} = 0 $$
However, I am having a hard time to prove the same result using the mean theorem for integrals. By the mean theorem we know that for each n, there exists $ c_{n} \in [0, 1] $ such that $ I_{n} = f(c_{n}) $ where $f:[0, \infty) \to [0, 1)$ is defined as $$ f(x) = \frac{x^{n}}{x^{n}+1}$$
Furthermore actually each such $ c_{n}$ cannot be either 0 or 1, thus $ c_{n} \in (0,1) \hspace{1mm} \forall n$. Indeed this follows from the below reasoning:
It can be easily verified that $ (I_{n})_{n\geq1} $ is a decreasing sequence and since $$ I_{1} = 1 - \ln(2) < \frac{1}{3}$$ it follows that $ I_{n} < \frac{1}{3} \forall n \geq 1$. Suppose there exists an $ n = n_{0}$ for which $ c_{n_{0}} = 1 $, then $$ I_{n_{0}} = f(c_{n_{0}}) = f(1) = \frac{1}{2} $$ which contradicts the above condition.
Suppose now that there exists an $ n = n_{1}$ for which $ c_{n_{1}} = 0 $. Then it follows that $$ I_{n_{1}} = f(c_{n_{1}}) = f(0) = 0 $$ But this cannot occur since $$ I_{n} = \int_{0}^{1}\frac{x^{n}}{x^{n}+1}dx \geq \int_{0}^{1}\frac{x^{n}}{2}dx = \frac{1}{2(n+1)} > 0 \quad \forall n \geq 1$$
Therefore $ c_{n} \in (0,1) \forall n \geq 1 $. Let's also show that the sequence $ (c_{n})_{n \geq 1} $ is decreasing. Indeed, we have the following:
$$ I_{n} = f(c_{n}) \geq I_{n+1} = f(c_{n+1}) $$ But f is a strictly increasing function on $ (0,1) $ which can be easily verified by computing its derivative. Therefore it must hold that $ c_{n} \geq c_{n+1} \forall n \geq 1$ hence $ (c_{n})_{n \geq 1} $ is decreasing and since it is also bounded in $ (0, 1) $, it must be convergent via the monotone convergence theorem.
Let $$ \lim_{n \to \infty} c_{n} = l$$ I think it would then be enough to show that this limit cannot be 1 so that we could conclude that $ I_{n} $ converges to 0. Indeed, if $ l< 1$ then we must have $ l \in [0, 1)$ since $ c_{n} \in (0, 1) \forall n \geq 1$ and thus we have: $$ \lim_{n \to \infty} I_{n} = \lim_{n \to \infty} f(c_{n}) = f(\lim_{n \to \infty} c_{n}) = f(l) = \frac{l^{n}}{l^{n}+1} $$ and the last term in the RHS above converges to 0 when $ n $ goes to infinity.
I don't know whether or not $ l $ can also be equal to 1 and also if my reasoning above for the conclusion that the desired limit is 0 for $ l \in [0, 1) $ is entirely correct. Could someone help me regarding this? Thank you so much!
As commented above by @jwhite, if $(c_n)$ was decreasing, you wouldn't need to prove that $l<1$. Anyway, your attempt is flawed because you are acting (twice) as if $f$ was a fixed function, not depending on $n$.
If, instead, you write $$\forall x\in[0,1],\quad f_{\color{red}n}(x):=\frac{x^n}{1+x^n},$$ you will realize that: