Answer
a) The limit provided in part(a) is not correct.
b) The limit provided in part(b) is correct.
Work Step by Step
(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, $\dfrac{0}{6}=0$
