Limit at positive infinity of a piecewise rational function

Since we only care about $x \to +\infty$, and the definition $g(x) = \frac{3x+2}{x+1}$ applies for all $x \ge 5$, the piece for $x<5$ is irrelevant here. So we need $\lim_{x\to+\infty} \frac{3x+2}{x+1}$.

Divide numerator and denominator by $x$ (the highest power of $x$ appearing in the denominator): $\frac{3x+2}{x+1} = \frac{3 + 2/x}{1 + 1/x}.$ As $x \to +\infty$, the terms $2/x$ and $1/x$ both go to 0, leaving $\frac{3+0}{1+0} = 3$.

Therefore $\lim_{x\to+\infty} g(x) = 3$, and since the numerator and denominator have equal degree, this matches the shortcut of taking the ratio of leading coefficients ($3/1$). This means $y = 3$ is a horizontal asymptote of $g$.