# Is $\displaystyle\lim_{h \to 0} H_n(f(h), g(h)) = H_n(\displaystyle\lim_{h \to 0} f(h), \displaystyle\lim_{h \to 0} g(h))$ true for all $n$?

Consider the limit $\displaystyle\lim_{h \to 0} H_n(f(h), g(h)),$ where $H_n(a, b)$ denotes the $n$th hyperoperation $H_n(a,b) = a \uparrow^{n-2}b$ with both $f(x)$ and $g(x)$ being continuous and well-defined for every complex number $z = a + bi.$

We know that this limit is equal to $H_n(\displaystyle\lim_{h \to 0} f(h), \displaystyle\lim_{h \to 0} g(h))$ if $n = 1$, that is $\displaystyle\lim_{h \to 0} \left(f(h) + g(h)\right) = \displaystyle\lim_{h \to 0} f(h) + \displaystyle\lim_{h \to 0} g(h)$.

We also know that this holds for $n = 2$; we can write $\displaystyle\lim_{h \to 0} \left(f(h) \cdot g(h)\right)$ as $\displaystyle\lim_{h \to 0} f(h) \cdot \displaystyle\lim_{h \to 0} g(h)$.

Now, this made me wonder if $\displaystyle\lim_{h \to 0} H_n(f(h), g(h)) = H_n(\displaystyle\lim_{h \to 0} f(h), \displaystyle\lim_{h \to 0} g(h))$ true for all $n \geq 0$? If not, are there any conditions that needs to be satisfied for the equality to hold? How does one go to prove a such thing?

If you modify you requirement to allow g(x) be a function to $\mathbb{N}$ then you can make use of the fact that $g(0) = \lim_{x \to 0} g(x)$ for any $g : S \to \mathbb{N}$. At that point the problem becomes trival.