Limit at infinity involving an exponential term

As $x \to +\infty$, the term $3^{-x} = \left(\frac{1}{3}\right)^x$ approaches $0$, since raising a number between $0$ and $1$ to larger and larger powers drives it toward zero. So both the numerator $5 + 3^{-x}$ and the denominator $2 - 3^{-x}$ approach finite, well-defined values as $x$ grows.

Substituting this limiting behavior gives $\frac{5 + 0}{2 - 0} = \frac{5}{2}$. Since the limit exists and equals a finite number, $g(x)$ has a horizontal asymptote at $y = \frac{5}{2}$ as $x \to +\infty$.