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

$$k(x)=1-\frac{1}{x-1}$$ The graph is shown below.
(What I present here is just an example. There are other possible functions which satisfy the exercise.) - As $\lim_{x\to1^-}k(x)=\infty$ and $\lim_{x\to1^+}k(x)=-\infty$, we would suppose the function $k(x)$ to be in the form of a rational function $k(x)=A/B$ where $B=x-1$ but $A\lt0$ so that we would have the reverse $\lim_{x\to1^-}k(x)=\infty$ and $\lim_{x\to1^+}k(x)=-\infty$. $$k(x)=\frac{A}{x-1}$$ - As $\lim_{x\to\pm\infty}k(x)=1$ $$\lim_{x\to\pm\infty}\frac{A}{x-1}=1$$ We can set $\lim_{x\to\pm\infty}\frac{A}{x-1}=0$ then add $1$ following. In detail, $$\lim_{x\to\pm\infty}\frac{A}{x-1}+1=1$$ Any negative numbers would suit $A$ now. I would pick $A=-1$ In conclusion, the function I come up with is $$k(x)=1-\frac{1}{x-1}$$ The graph is shown below.