How do I prove that $3x^3 +2x + 1 $ is $\omega(x \cdot \log x) $ I am trying to answer this question:
$3x^3 +2x + 1$ is $ \omega(x \cdot \log x)$

My question is how to solve this question.

Here is what I have tried so far:

I applied the definition $3x^3 + 2x + 1 &gt; c \cdot x\log x$ for all c, given $n \geq k$, for some k

I tried to apply the technique of applying the equality that:
$n &lt; n^3$

Since n is less than the highest order term we can just move all terms to order n ??

Giving $6x &lt; c \cdot x \log x$

Then you can choose:
$k &lt; 2^\frac{6}{c}$
The easiest way is to check that $\lim_{x \to \infty} \frac{3x^3 + 2x +1}{ x \log x} = +\infty$, which is a sufficient condition for $3x^3 + 2x +1 \in \omega(x \log x)$.

$$
\lim_{x \to \infty} \frac{3x^3 + 2x +1}{ x \log x} =
\lim_{x \to \infty} \frac{3x^3}{ x \log x} = 
\lim_{x \to \infty} \frac{3x^2}{  \log x} =
\lim_{x \to \infty}  6x^2 
= +\infty.
$$