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

$$g(x)=\frac{2}{x-3}$$
(What I present here is just an example. There are other possible functions which satisfy the exercise.) - As $\lim_{x\to3^-}g(x)=-\infty$ and $\lim_{x\to3^+}g(x)=\infty$, I would suppose that my function would have this rational form $A/B$, with $A\gt0$ and $B=x-3$ to make the function approach $\infty$ as $x\to3^+$ and approach $-\infty$ as $x\to3^-$. . $$g(x)=\frac{A}{x-3}$$ - As $\lim_{x\to\pm\infty}g(x)=0$: $$\lim_{x\pm\infty}\frac{A}{x-3}=0$$ Remember the method that divides both numerator and denominator by the highest degree of $x$ in the denominator, which is $x$ in this case: $$\lim_{x\pm\infty}\frac{\frac{A}{x}}{1-\frac{3}{x}}=0$$ $$\frac{\lim_{x\pm\infty}\frac{A}{x}}{1-0}=\lim_{x\pm\infty}\frac{A}{x}=0$$ We know that as $A\in Z$, $\lim_{x\to\pm\infty}(A/x)=0$, so we can choose any positive number $(A\gt0)$ here. I would choose $A=2$. So in conclusion, the function I come up with is $$g(x)=\frac{2}{x-3}$$. The graph is shown below.