If we have a normal distribution with mean $\mu=f(x)$ and variance $\sigma^2$ where $\sigma=g(x)$ having the form:
\begin{equation} p(y)=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(y-\mu)^2}{2\sigma^2}} \end{equation} along with the following conditions \begin{equation} \lim_{x\to\infty}\frac{f(x)}{g(x)}=K \\ \lim_{x\to\infty}f(x)=\infty \\ \lim_{x\to\infty}g(x)=\infty \end{equation} where $K$ is just a constant real number by taking the following limit \begin{equation} \lim_{x\to\infty}\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(y-\mu)^2}{2\sigma^2}} \end{equation}
what kind of distribution will I end up?
is this still a Normal? is there a way to represent it?
You're talking about limits, so to re-frame your question, as $x\rightarrow\infty$ does the distribution $\mathcal{N}(f(x),g(x)^2)$ converge to another distribution on $\mathbb{R}$? The weakest converge we usually deal with is convergence in distribution; if these distributions converge then their CDFs $F_x$ converge to a CDF $F$ wherever $F$ is continuous. As $F_x(y)=\Phi\left(\frac{y-f(x)}{g(x)}\right)\rightarrow\Phi(-K)$ as $x\rightarrow\infty$. So $F$ needs to be constant and equal to $\Phi(-K)$ everywhere. As CDFs can't be constant, these distributions don't converge.