Limit of this expression $ \lim \limits_{n\to \infty }\left( \frac{15n+7}{n \log\left( n\right) } \right) $ The limit $$ \lim \limits_{n\to \infty }\left( \frac{15n+7}{n \log\left( n\right) } \right) $$
Is it valid to do

$\frac{15 + 7/n}{\log n}$

which makes the numerator go to $15$ as the bottom goes to infinity, therefore the whole thing goes to $0$? How would I show more formally that this is true?
Holds the following
$$0\leq\frac{15n + 7}{n\log n}\leq \frac{15n + 7n}{n\log{n}}\leq\frac{22}{\log{n}}&lt;\varepsilon$$ for $n&gt;e^\frac{22}{\varepsilon}$ and arbitrary $\varepsilon$.

By letting $\varepsilon\to 0$ and squeeze theorem, we can conclude $$\lim_{n\to\infty}\frac{15n + 7}{n\log n}=0.$$
Yes it is fully valid that $\lim_{n \to \infty} \frac{15n +7}{n \log n} = \lim_{n \to \infty} \frac{15 +\frac{7}{n}}{ \log n}$, and hence the limit is $0$.  

To prove this, note that $\lim_{n \to \infty} \frac{\frac{1}{n}}{\frac{1}{n}} =1$. So, 

$\lim_{n \to \infty} \frac{15n +7}{n \log n} = \lim_{n \to \infty} \frac{\frac{1}{n}(15n +7)}{\frac{1}{n}(n \log n)} =  \lim_{n \to \infty} \frac{15 +\frac{7}{n}}{ \log n} = 0$.