what is the limit of $\underset{x\to 0}{\mathop{\lim }}\,{{\left( \int_{0}^{x}{{{e}^{{{t}^{2}}}}dt} \right)}^{1/x}}$ I tired first to suppose that $I=\underset{x\to 0}{\mathop{\lim }}\,{{\left( \int_{0}^{x}{{{e}^{{{t}^{2}}}}dt} \right)}^{1/x}}$ then take both sides as $\ln$ to get ,
$\ln I =\underset{x\to 0}{\mathop{\lim }}\frac{1}{x}\ln\int_{0}^{x}{{{e}^{{{t}^{2}}}}dt}$
Can this road leads us to the solution or we need to use another shortcut!
 A: Since $e^{t^2}$ is increasing you have $$0 < \int_0^x e^{t^2} \, dt < x e^{x^2}$$ and thus $$0 < \left( \int_0^x e^{t^2} \, dt \right)^{1/x}  < x^{1/x} e^x$$ for all $x > 0$. Since $\dfrac{\ln x}x \to -\infty$ as $x \to 0^+$ you find that $x^{1/x} \to 0$ as $x \to 0^+$.  Now apply the squeeze theorem.
A: Knowing the Gaussian integral, one can use the Sandwich theorem as follows:
$$\color{red}0 \le \color{blue}{\left(\int_0^x e^{t^2}\mathrm dt\right)^{1/x}}\le \left(\int_0^{\infty}e^{t^2}\mathrm dt\right)^{1/x} = \left(\frac{\sqrt{\pi}}{2}\right)^{1/x}\to \color{red}0$$
Therefore,
$$\lim_{x\to0}\color{blue}{\left(\int_0^x e^{t^2}\mathrm dt\right)^{1/x}} = \color{red}0$$
A: As you showed,
$$\ln{I} = \lim_{x\to0}\Bigg(\frac{1}{x}\cdot\ln\bigg(\int_{0}^{x}e^{t^2}dt\bigg)\Bigg)$$
This allows you to deduce the limit immediately. The limit of $\frac{1}{x}$ as $x\to0^{+} $ is $\infty$. Meanwhile, the limit of $\ln{\int}$ is $-\infty$. Thus,
$$\ln{I} \to -\infty$$
as $x \to 0^+$.
$$\implies I \to 0$$
as $x \to 0^+$.
