# Doubts regarding limits and logarithms Lets say you are given this limit lim_(n->oo) ( log(n + n^n + n^(1/n) )

Doubts regarding limits and logarithms
Lets say you are given this limit
$\underset{n\to \mathrm{\infty }}{lim}\left(\mathrm{log}\left(n+{n}^{n}+{n}^{1/n}\right)$
That expression is equal to
$\mathrm{log}\left(\underset{n\to \mathrm{\infty }}{lim}\left[n+{n}^{n}+{n}^{1/n}\right]\right)$
isn't it?
My question is if I could descompose the limit like this without changing the limit like this
$\mathrm{log}\left(\underset{n\to \mathrm{\infty }}{lim}n+\underset{n\to \mathrm{\infty }}{lim}{n}^{n}+\underset{n\to \mathrm{\infty }}{lim}{n}^{1/n}\right)$
Could I?
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Claire Love
First of all, a sufficient condition to interchange the limit and a function is that the function is continuous. So for $f\left(x\right)$ continuous we have
$\underset{n\to \mathrm{\infty }}{lim}f\left({a}_{n}\right)=f\left(\underset{n\to \mathrm{\infty }}{lim}{a}_{n}\right)$
since the logarithm is continuous, you can
$\underset{n\to \mathrm{\infty }}{lim}\mathrm{log}\left({a}_{n}\right)=\mathrm{log}\left(\underset{n\to \mathrm{\infty }}{lim}{a}_{n}\right)$
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
$\underset{n\to \mathrm{\infty }}{lim}\left({a}_{n}+{b}_{n}\right)=\underset{n\to \mathrm{\infty }}{lim}\left(\frac{1}{n}+\frac{3n+1}{7n-2}\right)=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}+\underset{n\to \mathrm{\infty }}{lim}\frac{3n+1}{7n-2}=0+\frac{3}{7}=\frac{3}{7}$
works, since both limits $\underset{n\to \mathrm{\infty }}{lim}{a}_{n},\underset{n\to \mathrm{\infty }}{lim}{b}_{n}$ do exist. But for example
$\underset{n\to \mathrm{\infty }}{lim}\left(0\right)=\underset{n\to \mathrm{\infty }}{lim}\left(n-n\right)\ne \underset{n\to \mathrm{\infty }}{lim}n-\underset{n\to \mathrm{\infty }}{lim}n$
obviously doesn't work since the limit of $\underset{n\to \mathrm{\infty }}{lim}n$ does not exist. In your case
$\underset{n\to \mathrm{\infty }}{lim}\left(\mathrm{log}\left(n+{n}^{n}+{n}^{1/n}\right)\ne \mathrm{log}\left(\underset{n\to \mathrm{\infty }}{lim}n+\underset{n\to \mathrm{\infty }}{lim}{n}^{n}+\underset{n\to \mathrm{\infty }}{lim}{n}^{1/n}\right)$
because $\underset{n\to \mathrm{\infty }}{lim}{n}^{n}$ and $\underset{n\to \mathrm{\infty }}{lim}n$ don't exist.