We have to find $\lim_{n\rightarrow \infty}\int_0^1 e^{x^n}$ dx and also justify the existence.
Answer is 1. Writing it rigorously (to justify existence) is a little problem for me. I have tried through uniform convergence and attached.
My problem is that the proof which I have written may not be right as the N is dependent on 'b' which should not be the case in uniform convergence. Kindly help suggesting a better proof.
On
Clearly $e^{x^n} \ge 1$, hence the integral is $\ge 1$ for all $n$. On the other hand, we have for each $\delta>0$ $$ \int_{0}^1e^{x^n}\mathrm{d}x\le \int_{0}^{1-\delta}e^{x^n}\mathrm{d}x+\delta e \le e^{(1-\delta)^n}+\delta e. $$
Now, $(1-\delta)^n<\delta$ if $n$ is sufficiently large, then $$ \limsup_{n\to \infty}\int_{0}^1e^{x^n}\mathrm{d}x\le \inf_{\delta>0}\limsup_{n\to \infty}\left(e^{(1-\delta)^n}+\delta e\right)\le \inf_{\delta>0}(e^{\delta}+3\delta)\le 1. $$
On
All we need is the inequality $e^u\le 1+3u, u \in [0,1].$ (To prove it, note they agree at $u=0$ and the derivative of the right side is $3,$ which is greater than the derivative of $e^u$ on this interval.)
Thus
$$1 \le \int_0^1e^{x^n}\, dx \le \int_0^1(1+3x^n)\, dx = 1 + 3/(n+1).$$
The limit is therefore $1$ by the squeeze theorem.
To show $\lim\limits_{n\rightarrow\infty}{\int\limits_{0}^{1}{e^{x^n}\,dx}} = 1$, it suffices to show that $\lim\limits_{n\rightarrow\infty}{\int\limits_{0}^{1}{(e^{x^n}-1)\,dx}} = 0$. Clearly, we have $\int\limits_{0}^{1}{(e^{x^n}-1)\,dx}\ge 0$. On the other hand, for any $0\le y\le 1$, we have $$e^y-1 = \int\limits_{0}^{y}{e^t\,dt}\le \int\limits_{0}^{y}{e\,dt} = ey$$ since, for $0\le t\le y\le 1$, we have $e^t\le e$. It follows that $e^{x^n}-1\le ex^n$ for all $0\le x\le 1$ for any $n$, and so $$ \int\limits_{0}^{1}{(e^{x^n}-1)\,dx} \le \int\limits_{0}^{1}{ex^n\,dx} = \frac{e}{n+1}.$$ It follows that $$ 0\le \int\limits_{0}^{1}{(e^{x^n}-1)\,dx} \le \frac{e}{n+1} $$ and so by the Squeeze theorem we have $\lim\limits_{n\rightarrow\infty}{\int\limits_{0}^{1}{(e^{x^n}-1)\,dx}} = 0$, as desired.