A function $f: \mathbb{R}\to \mathbb{R}$ is an infinitesimal for $a$ if $\lim_{x\to a}f(x) = 0$. Two infinitesimals $f$ and $g$ are equivalent if $\lim_{x\to a} \dfrac{f(x)}{g(x)} = 1$. It is know n that an infinitesimal can be replaced by an equivalent one when we have products and sums (but not sums or differences). It is easy to shown that this also works inside $n^{th}$-roots. That is, if $\lim_{x\to a} \dfrac{f(x)}{g(x)} = 1$, then $\lim_{x\to a} \dfrac{f(x)^{1/n}}{g(x)^{1/n}} = 1$.
My question is, can this be generalized to the case of continuous functions, or certain families of continuous functions?
I am trying to find $\lim_{x\to 0} (\cos(x)^2)^{1/sin^2(x)}$ without using L´Hôpital, and I was wondering if I could do the substitutions of $\sin(x)$ and $\cos(x)$ after applying logarithms. The result is the same as solving the limit by L´Hôpital, so there is a chance this could work.
If it is not the case, I'd appreciate some guidance to solve the limit.
Edit: I also though that, if we define $y = 1/\sin(x)^2$, then $x\to 0$ implies that $y\to \infty$, and, since $\cos(x)^2 = 1-\sin(x)^2$, the original limit becomes $$ \lim_{y\to \infty} (1-\frac{1}{y})^y = \frac{1}{e} $$
Is this correct?