Limit of an integral

Relevant wiki: Fundamental Theorem of Calculus

$\displaystyle \large \begin{aligned} \displaystyle L&=\lim_{x\to\infty} \dfrac{\int\limits_0^x e^{t^2}\; dt}{e^{x^2}}\quad\quad\text{L-Hopital's Form}\\ &=\lim_{x\to\infty}\dfrac{e^{x^2}}{2x\times e^{x^2}} \\ &=\lim_{x\to\infty}\dfrac{1}{2x} \\ &=0\end{aligned}$

Relevant wiki: Squeeze Theorem

$Since$ $0 \leq t \leq x$ $, \:then$ $t^2 \leq tx$ $,\: thus$ $e^{t^2}\leq e^{tx}$ $. \: It\: follows\: that$ $\int_0^x e^{t^2}dt\leq\int_0^x e^{tx}dt=\frac{e^{x^2}-1}{x}$ .

$So,$ $e^{-{x^2}}\int_0^x e^{t^2}dt\leq e^{-{x^2}}\int_0^x e^{tx}dt=e^{-{x^2}}(\frac{e^{x^2}-1}{x})=\frac{1}{x}(1-e^{-{x^2}})$ .

$Hence,$ $0\leq\lim_{x \to ∞}e^{-{x^2}}\int_0^x e^{t^2}dt\leq\lim_{x \to ∞}\frac{1}{x}(1-e^{-{x^2}})=0(1-0)=0$ .