I have been given the sequence $x_n=\frac{\log(3n+2)}{\log(n^2+2)}$ which has the limit $\frac{1}{2}$, and I have been tasked with proving that.
This is the first time I have ever proven a sequence, and I apologize for any glaring mistakes.
Thus, to prove convergence, I have to show that for any $\varepsilon$> 0 there exists an $N$ such that for all $n > N$ the distance between $x_n$ and $\frac{1}{2}$ is less than $\varepsilon$.
$$\left|x_n - \frac{1}{2}\right| = \left| \frac{\log(3n+2)}{\log(n^2+2)} - \frac{1}{2}\right| < \varepsilon $$
I multiply both sides with $\log(n^2+2)$ and get :$$\left|\log(3n+2)-\frac{\log(n^2+2}{2}\right| < \varepsilon \log(n^2+2) $$
I now use the property of logarithms:$$\left|\frac{3n+2}{(n^2+2)^\frac{1}{2}}\right| < e^{\varepsilon \cdot \log(n^2+2)}$$Which I simplify to:$$\left|\frac{3n+2}{(n^2+2)^\frac{1}{2}}\right| < (n^2+2)^\varepsilon $$Thus, I choose $N$ such that $(n^2+2)^\varepsilon$< 1 +$\frac{\varepsilon}{2}$ for all $n≥N$, and for all $n≥N$ the distance between $x_n$ and $\frac{1}{2}$ is less than $\varepsilon$.