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?
 A: Holds the following
$$0\leq\frac{15n + 7}{n\log n}\leq \frac{15n + 7n}{n\log{n}}\leq\frac{22}{\log{n}}<\varepsilon$$ for $n>e^\frac{22}{\varepsilon}$ and arbitrary $\varepsilon$.
By letting $\varepsilon\to 0$ and squeeze theorem, we can conclude $$\lim_{n\to\infty}\frac{15n + 7}{n\log n}=0.$$
A: Your reasoning is perfectly fine, and can be made fully rigorous as follows.
$\def\nn{\mathbb{N}}$
$\def\rr{\mathbb{R}}$

$\lim_{n\to\infty} \dfrac{15n+7}{n \ln(n)}$
$\ = \lim_{n\to\infty} \dfrac{15+7/n}{\ln(n)}$   [the division is valid because $n \ne 0$ and it does not change the value]
$\ = \dfrac{\lim_{n\to\infty}( 15+7/n )}{\lim_{n\to\infty} \ln(n)}$   [since these two limits exist in $\overline{\rr}$ and the ratio is defined in $\overline{\rr}$]
$\ = \dfrac{\lim_{n\to\infty} 15 + \lim_{n\to\infty} (7/n) }{\infty}$   [since these two limits exist in $\overline{\rr}$ and the sum is defined in $\overline{\rr}$]
$\ = \dfrac{15+0}{\infty} = 0$.

Note that this reasoning is carried out using the affinely extended reals $\overline{\rr}$, which includes $\infty$ and $-\infty$ but where certain operations are not allowed. The usual limit laws can be proven to work here as well as long as the extra restrictions as stated above are met.
Also, your question about validity of division by $n$ is a good one. Technically one has to specify the domain over which the limit is taken. I've never seen a textbook that specifies it, but it's important! For instance is $n$ a real number? Is it positive? In this case I assume you want $n \in \nn$ but you still need $n \ne 1$ otherwise $n \ln(n)$ would be zero. Personally I would write "$\lim_{n \in \nn_{>1} \to \infty} \dfrac{15n+7}{n \ln(n)}$" but I've never seen anyone else do that (they all rely on the readers to figure out the context...).
A: Your reasoning is fine but when you write it as an answer it needs to be made precise. Thus we have $$\lim_{n \to \infty}\frac{15n + 7}{n\log n} = \lim_{n \to \infty}\frac{15 + 7/n}{\log n} = \lim_{n \to \infty}(15 + 7/n)\cdot\frac{1}{\log n} = (15 + 0)\cdot 0 = 0$$ Often when a function $f(n) \to \infty$ as $n \to \infty$ then it is better to say that $1/f(n) \to 0$ as $n \to \infty$ and then we can use algebra of limits directly as I have shown above.
A: 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, 
$\lim_{n \to \infty} \frac{15n +7}{n \log n} = \lim_{n \to \infty} \frac{\frac{1}{n}(15n +7)}{\frac{1}{n}(n \log n)} =  \lim_{n \to \infty} \frac{15 +\frac{7}{n}}{ \log n} = 0$.
A: $$\lim_{n\to \infty} \log n=\infty$$
That is enough to show that
$$\lim_{n\to\infty}\frac{15n + 7}{n\log n}=0.$$
They are the key limits
$$\lim_{n\to \infty} \log n=\infty$$
$$\lim_{n\to \infty} \frac{n^{m_1}}{n^{m_2}}=
\begin{cases}
0\quad, m_1<m_2\\
\infty\quad, m_1>m_2
\end{cases}
$$
$$\lim_{n\to \infty} \frac{\log n}{n}=0$$
Always remember that the limit of logarithm is infinity but it has no chance to overcome polynomials.
