$\lim_{x\to 0}(\frac{3x^2+2}{5x^2+2})^{\frac{3x^2+8}{x^2}}$ using Taylor series expansion or any other expansion Can we evaluate this $\lim_{x\to 0}(\frac{3x^2+2}{5x^2+2})^{\frac{3x^2+8}{x^2}}$ using Taylor/Maclaurin series by expanding the function about $x=0?$
I can otherwise solve this limit.This is in the form of $1^{\infty}$.
$$\lim_{x\to 0}\left(\frac{3x^2+2}{5x^2+2}\right)^{\frac{3x^2+8}{x^2}}=\exp \left[{\lim_{x\to 0}\left(\frac{3x^2+2}{5x^2+2}-1\right)\times \frac{3x^2+8}{x^2}}\right]=e^{-8}$$
But i want to know,cant it be found using Taylor/Maclaurin series?What are the limitations of Taylor/Maclaurin series?
When can we not use them?
Is it true that we can use them whenever L Hospital rule is applicable(in $\frac{0}{0}$ and $\frac{\infty}{\infty}$ cases.)Can we use them in $0^0,\infty^0,1^\infty$ cases.
Please help me.Thanks.

Notice, we know $\color{blue}{\lim_{n\to \infty}\left(1-\frac{1}{n}\right)^n=e^{-1}}$ 
Now we have 
$$\lim_{x\to 0}\left(\frac{3x^2+2}{5x^2+2}\right)^{\Large \frac{3x^2+8}{x^2}}$$$$=\lim_{x\to 0}\left(1-\frac{2x^2}{5x^2+2}\right)^{\Large \frac{3x^2+8}{x^2}}$$
let, $\large \frac{2x^2}{5x^2+2}=\frac{1}{t}$$\implies x^2=\large \frac{2}{2t-5}$, hence
$$=\lim_{t\to \infty}\left(1-\frac{1}{t}\right)^{8t-17}$$
$$=\lim_{t\to \infty}\left(1-\frac{1}{t}\right)^{8t}\left(1-\frac{1}{t}\right)^{-17}$$
$$=\left(\lim_{t\to \infty}\left(1-\frac{1}{t}\right)^t\right)^{8}\cdot \lim_{t \to \infty}\left(1-\frac{1}{t}\right)^{-17}$$
$$=(e^{-1})^{8}\cdot (1)^{-17}=\color{red}{\large e^{-8}}$$
