Answer
$$\lim_{x\to3^+}\frac{1}{x-3}=\infty$$
Work Step by Step
$$A=\lim_{x\to3^+}\frac{1}{x-3}$$
As $x\to3^+$, $x-3$ approaches $0$ from the right, where $(x-3)\gt 0$, so $1/(x-3)$ will approach $\infty$. Therefore,
$$A=\lim_{x\to3^+}\frac{1}{x-3}=\infty$$
