We are required to calculate $\int^{\infty}_{0}e^{-x^2}dx$ in three steps:
1) Show $\forall t \in \mathbb{R}, 1-t\le e^{-t}$ and $1+t \le e^t$ (This is easily proven; no need to elaborate on this)
2)Show $\forall n \in \mathbb{N}: \int^{\sqrt{n}}_{0}(1-\frac{x^2}{n})^{n}dx\le\int^{\infty}_{0}e^{-x^2}dx\le\int^{\infty}_{0}\big(1+\frac{x^2}{n}\big)^{-n}dx$
Comment: This is where I start having problems. my initial thought was to look at $\big(1-\frac{x^{2}}{n}\big)^{n}$ and we know $\big(1-\frac{x^2}{n}\big)^n\to e^{-x^2}$ and $\big(1-\frac{x^2}{n}\big)^{n}$ is monotone increasing which means $\big(1-\frac{x^2}{n}\big)^n \le e^{-x^2} \forall n\in\mathbb{N}$. Is this correct and I'm not certain on the second inequality.
3)Finally, we should use $\forall n \in \mathbb{N}: \int^{\sqrt{n}}_{0}\big(1-\frac{x^2}{n}\big)^ndx\leq\int^{\infty}_{0}e^{-x^2}dx\le \int^{\infty}_{0}\big(1+\frac{x^2}{n}\big)^{-n}dx$ as well as the substitutions $x=\sqrt{n}\sin t$ (I presume for the first integral) and $x=\sqrt{n}\tan t$ (I presume for the third integral whereby $ t\in \big[0,\frac{\pi}{2}\big]$. I am struggling to integrate the first and third integral because both still have terms in $n$. How can I do this?
Help is greatly appreciated.
Hint
For the second one you obtain: $$\sqrt{n}\int_0^\frac{\pi}{2} \left(\frac{1}{\cos^2(t)}\right)^{-n} \frac{1}{\cos^2(t)}dt=\sqrt{n}\int_0^\frac{\pi}{2} \cos(t)^{2n-2} dt$$ and again from the equivalent of Wallis' integrals you obtain: $$\sqrt{n}\int_0^\frac{\pi}{2} \cos(t)^{2n-2} dt \sim \frac{\sqrt{\pi}}{2}$$