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 states that $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{A'(x)}{B'(x)}$
Here, $\lim\limits_{x \to 3} \dfrac{x-3}{x^2-3}=\lim\limits_{x \to 3} \dfrac{1}{2x}=\dfrac{1}{6}$
Also, $f(3)=\dfrac{3-3}{3^2-3}=\dfrac{0}{6} \ne \dfrac{0}{0}$
(L'Hospital's rule can not be used because we do not have indeterminate form.)
b) $\lim\limits_{x \to 3} \dfrac{x-3}{x^2-3}=\dfrac{3-3}{3^2-3}=\dfrac{0}{6}=0$
Hence,
a) The limit provided in part(a) is not correct.
b) The limit provided in part(b) is correct.