Let $l,n \geq 1$ two integers, and consider both
$$ f_{l,n}(x) = \begin{cases} 1 - \frac{n(l+1)l}{6}x^2 & |x| \leq \sqrt{\frac{6}{n(l+1)l}} \\ 0 & \text{otherwise} \end{cases} $$
and
$$ g_{l,n}(x) = e^{-\frac{n\left(l+1\right)l}{6} x^2} $$
Is it true that $f_{l,n}(x) \sim g_{l,n}(x)$ when $n \to +\infty$?
My answer would be yes, because $\left\lVert f_{l,n} \right\rVert_{\infty},\left\lVert g_{l,n} \right\rVert_{\infty} \leq 1$, and $f_{l,n}(0) = g_{l,n}(0) = 1$, otherwise we have $0 \leq f_{l,n}(x), g_{l,n}(x) < 1$ therefore when $n$ diverges the limits are the same, which yields $f_{l,n} \sim g_{l,n}$. Is this argument correct?
It is true that $\lim_{n\to\infty}\bigl(\,f_{l,n}(x)-g_{l,n}(x)\bigr)=0$ for all $x\in\Bbb R$. But $$ f_{l,n}\Bigl(\sqrt{\frac{6}{n(l+1)l}}\Bigr)=0\quad \text{while}\quad g_{l,n}\Bigl(\sqrt{\frac{6}{n(l+1)l}}\Bigr)=e^{-1}. $$ Thus $$ \|f_{l,n}-g_{l,n}\|_\infty\ge e^{-1}. $$ So it depends on what you mean by $\sim$. Let's compute the quotient: $$ \frac{f_{l,n}(x)}{g_{l,n}(x)}=\begin{cases} e^{\tfrac{n\left(l+1\right)l}{6} x^2}\Bigl(1 - \dfrac{n(l+1)l}{6}x^2\Bigr) & \text{if }|x| \le \sqrt{\dfrac{6}{n(l+1)l}}\ ,\\ 0 & \text{otherwise.} \end{cases} $$ If $x\ne0$, this converges to $0$ as $n\to\infty$. If $x=0$, the quotient is $1$.
Observe that $\frac{g_{l,n}(x)}{f_{l,n}(x)}$ is not defined.