Proving little o by first principles I'm trying to prove that $15n+7$ is $o(n\log n)$ (by first principles, i.e. no limits).

My idea is to solve for n to determine $n_0$ and then work backwards from there. But I can't seem to find a way.

$15n+7 &lt; cn\log n$

$7 &lt; n(c\log n-15)$

I got stuck here.
The problem is to prove that
$$
\bigg( \frac{15n + 7}{n\log n} \bigg) \to 0
$$
as $n \to \infty$;
for then we have by definition $15n + 7$ is $o(n\log n)$.

If $n &gt; 1$, then
$$
\frac{15n + 7}{n\log n} = \frac{15}{\log n} + \frac{7}{n\log n} &lt; \frac{15}{\log n} + \frac{7}{(\log n)^{2}}.
$$
Let $\varepsilon &gt; 0$. Note that we have 
$$
\frac{15}{\log n} &lt; \frac{\varepsilon}{2}
$$
if $n &gt; e^{30/\varepsilon} =: n_{1}$ and that we have
$$
\frac{7}{(\log n)^{2}} &lt; \frac{\varepsilon}{2}
$$
if $n &gt; e^{\sqrt{14/\varepsilon}} =: n_{2}$.
So taking $n_{0} := \max \{ 1, n_{1}, n_{2} \}$ suffices.
If $15n+7$ is $o(n \log n)$ then 

$$15n+7 &lt; cn \log_2n$$

for every $n \geq n_0$ for some positive real $c$ and $n_0$.

$$7 &lt; n(c\log_2n - 15)$$

If we chose $n_0 = 1$ then this would change to $7 &lt; -15$ which is false, so we need a larger $n$. Let's do $n_0 \geq 2$. 

$$7 &lt; 2(c - 15) = 2c - 30$$

And $7 &lt; 2c -30$ is true for $c &gt; 18.5$, so the proof is complete.