Lets say you are given this limit
$$\lim_{n\to\infty} ( \log(n + n^n + n^{1/n} )$$
That expression is equal to $$\log( \lim_{n\to\infty}[ n + n^n + n^{1/n}] ) $$ isn't it?
My question is if I could descompose the limit like this without changing the limit like this
$$\log( \lim_{n\to\infty} n + \lim_{n\to\infty} n^n + \lim_{n\to\infty} n^{1/n}) $$
Could I?
On
First of all, a sufficient condition to interchange the limit and a function is that the function is continuous. So for $f(x)$ continuous we have $$ \lim_{n\to\infty}f(a_n)=f(\lim_{n\to\infty}a_n) $$ since the logarithm is continuous, you can $$ \lim_{n\to\infty}\log(a_n)=\log(\lim_{n\to\infty}a_n) $$ interchange.
To answer the second part of your question: No in this particular case you cannot split the limit! For this to hold all limits need to exist on their own, for example
$$
\lim_{n\to\infty}(a_n+b_n)=\lim_{n\to\infty}(\frac 1 n +\frac {3n+1}{7n-2})=\lim_{n\to\infty}\frac 1 n+\lim_{n\to\infty}\frac {3n+1}{7n-2}=0+\frac 3 7=\frac 37
$$
works, since both limits $\lim_{n\to\infty}a_n,\lim_{n\to\infty}b_n$ do exist. But for example
$$
\lim_{n\to\infty}(0)=\lim_{n\to\infty}(n-n)\neq\lim_{n\to\infty}n-\lim_{n\to\infty}n
$$
obviously doesn't work since the limit of $\lim_{n\to\infty}n$ does not exist. In your case
$$
\lim_{n\to\infty} ( \log(n + n^n + n^{1/n} )\neq\log( \lim_{n\to\infty} n + \lim_{n\to\infty} n^n + \lim_{n\to\infty} n^{1/n})
$$
because $\lim_{n\to\infty} n^n$ and $\lim_{n\to\infty} n$ don't exist.
You can go from Limit ( log(n + n^n + n^(1/n) ) to log( limit n + n^n + n^(1/n) ) as log is continuous. Going from log( limit n + n^n + n^(1/n) ) to log( limit n + limit n^n + limit n^(1/n)) you can do as limits are linear. This all assumes that the relevant limits exists.