Answer
$$\lim_{x\to\infty}h(x)=\lim_{x\to-\infty}h(x)=\frac{3}{4}$$
Work Step by Step
$$h(x)=\frac{3-(2/x)}{4+(\sqrt2/x^2)}$$
(a) As $x\to\infty$, $x^2$ also approaches $\infty$ and therefore, both $\sqrt2/x^2$ and $2/x$ will approach $0$.
Therefore, $$\lim_{x\to\infty}h(x)=\lim_{x\to\infty}\frac{3-(2/x)}{4+(\sqrt2/x^2)}=\frac{3-0}{4+0}=\frac{3}{4}$$
(b) As $x\to-\infty$, $x^2$ will approach $\infty$ and therefore, both $\sqrt2/x^2$ and $2/x$ still approaches $0$.
Therefore, $$\lim_{x\to-\infty}h(x)=\lim_{x\to-\infty}\frac{3-(2/x)}{4+(\sqrt2/x^2)}=\frac{3-0}{4+0}=\frac{3}{4}$$
A graph of the function $h(x)$ is enclosed below, which shows that $h(x)$ approaches $3/4$ as $x$ approaches either $\infty$ or $-\infty$.
