## University Calculus: Early Transcendentals (3rd Edition)

$h(x)=1$ as $x\ge0$ and $h(x)=-1$ as $x\lt0$.
(What I present here is just an example. There are other possible functions which satisfy the exercise.) - As $\lim_{x\to0^-}h(x)=-1$ and $\lim_{x\to0^+}h(x)=1$, there is a break as $x$ moves from the left to the right of $0$. So I would create 2 functions for $h(x)$, one where $x\lt0$ and the other where $x\ge0$. - For $x\lt0$: $\lim_{x\to-\infty}h(x)=\lim_{x\to0^-}h(x)=-1$. A simple way to satisfy these is to make the function $h(x)=-1$ as $x\lt0$. - For $x\ge0$: $\lim_{x\to\infty}h(x)=\lim_{x\to0^+}h(x)=1$. A simple way to satisfy these is to make the function $h(x)=1$ as $x\ge0$. In conclusion, the function $h(x)=1$ as $x\ge0$ and $h(x)=-1$ as $x\lt0$. The graph is shown below.