Find $\lim \limits_{n \to \infty}{1*4*7*\dots(3n+1) \over 2*5*8* \dots (3n+2)}$

You are on the right track. Use the inequality $\ln(1+x)\ge x-x^2/2$ for $x\ge0$: $$ \sum_{k=0}^n\ln\Bigl(1+\frac1{3\,k+1}\Bigr)\ge\sum_{k=1}^n\Bigl(\frac{1}{3\,k+1}-\frac12\,\frac{1}{(3\,k+1)^2}\Bigr)=\sum_{k=1}^n\frac{1}{3\,k+1}-\frac12\sum_{k=1}^n\frac{1}{(3\,k+1)^2}. $$ The first sum goes to $+\infty$ because the series $\sum_{k=0}^\infty1/(3\,k+1)$ is divergent, and the second sum is bounded because the series $\sum_{k=0}^\infty1/(3\,k+1)^2$ is convergent. This means that the original limit is equal to $0$.