Fix an arbitrary $\varepsilon>0$. Since the function $y=e^{t}$ is convex, for $t \in[0, \varepsilon]$ its graph lies above the tangent $y=1+t$ but under the secant $y=1+a_{\varepsilon} t$, where $a_{\varepsilon}=\frac{1}{\varepsilon}\left(e^{\varepsilon}-1\right)$. Hence for $\frac{1}{n} \leq \varepsilon$ it holds  $$ \left(\int_{0}^{1} e^{x^{2} / n} d x\right)^{n} \geq\left(\int_{0}^{1}\left(1+\frac{x^{2}}{n}\right) d x\right)^{n}=\left(1+\frac{1}{3 n}\right)^{n} \rightarrow e^{1 / 3}, \text { as } n \rightarrow \infty, $$  and  $$ \left(\int_{0}^{1} e^{x^{2} / n} d x\right)^{n} \leq\left(\int_{0}^{1}\left(1+\frac{a_{\varepsilon} x^{2}}{n}\right) d x\right)^{n}=\left(1+\frac{a_{\varepsilon}}{3 n}\right)^{n} \rightarrow e^{a_{\varepsilon} / 3}, \text { as } n \rightarrow \infty $$  Since $a_{\varepsilon}=\frac{1}{\varepsilon}\left(e^{\varepsilon}-1\right) \rightarrow 1$, as $\varepsilon \rightarrow 0+$, we get $\lim _{n \rightarrow \infty}\left(\int_{0}^{1} e^{x^{2} / n} d x\right)^{n}=e^{1 / 3}$.  Answer: $e^{1 / 3}$
Find a limit  $$ \lim _{n \rightarrow \infty}\left(\int_{0}^{1} e^{x^{2} / n} d x\right)^{n}. $$