Limit at infinity via Squeeze Theorem

Since $-1 \le \sin(x) \le 1$ for all $x$, dividing by $x^2$ (which is positive for $x > 0$) gives $-\frac{1}{x^2} \le \frac{\sin(x)}{x^2} \le \frac{1}{x^2}.$ As $x \to +\infty$, both $-\frac{1}{x^2}$ and $\frac{1}{x^2}$ approach 0.

By the Squeeze Theorem, $\lim_{x\to+\infty} \frac{\sin(x)}{x^2} = 0$. Adding the constant 3 to all parts of the inequality and taking the limit gives $\lim_{x\to+\infty} h(x) = 0 + 3 = 3$. Since $h(x)$ approaches the finite value 3 as $x \to +\infty$, this confirms $y = 3$ is a horizontal asymptote of $h(x)$.