I am trying to calculate the limitation
$$ \lim_{x\to 0} = \frac{\exp({-(\frac{a}{2x}-c)^2})}{\frac{ax}{2}-cx^2}. $$ But by the l'Hôpital's rule, the limit remains to be 0/0 indetermined type. As $$\frac{a\exp({-(\frac{a}{2x}-c)^2})(\frac{a}{2x}-c)}{x^2(\frac{a}{2}-2cx)}$$.
How can I calculate a "$\frac{0}{0}$" $ $ t
On
We have that $\forall t\in \mathbb{R}$
$$e^{t^2}>t^2 \implies e^{-t^2}<\frac1{t^2}$$
therefore for $a\neq 0$
$$\left|\frac{e^{-\left(\frac{a}{2x}-c\right)^2}}{\frac{ax}{2}-cx^2}\right|<\frac{1}{\left|\frac{ax}{2}-cx^2\right|\left(\frac{a}{2x}-c\right)^2}=\frac{1}{\left|\frac{a^3}{8x}-\frac{a^2c}{2}+\frac{ac^2x}{2}-\frac{a^2c}{4}+ac^2x-c^3x^2\right|}\to 0$$
To simplify the computation, set $t:=\dfrac a{2x}-c$ and
$$\lim_{x\to 0}\frac{\exp\left({-\left(\dfrac{a}{2x}-c\right)^2}\right)}{\dfrac{ax}{2}-cx^2}=\lim_{t\to \infty}\frac{4(t+c)^2\exp\left(-t^2\right)}{a^2t}.$$
It is well known that an exponential increases faster than any polynomial and the limit is $0$.
You can decompose the factor in front of the exponential in terms of $t,1$ and $t^{-1}$. And for total confidence,
$$\lim_{t\to\infty}\frac t{e^{t^2}}=\lim_{t\to\infty}\frac 1{2te^{t^2}}=0.$$