## Thomas' Calculus 13th Edition

(a) L-Hospital's rule is defined as $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$ Here, $\lim\limits_{x \to 3} \dfrac{x-3}{x^2-3}=\lim\limits_{x \to 3} \dfrac{1}{2x}=\dfrac{1}{6}$ Also, $f(3)=\dfrac{3-3}{3^2-3}=\dfrac{0}{6} \ne \dfrac{0}{0}$ Here, we cannot use L-Hospital's rule because we do not have any indeterminate form (b) $\lim\limits_{x \to 3} \dfrac{x-3}{x^2-3}=\dfrac{3-3}{3^2-3}$ or, $\dfrac{0}{6}=0$