Evaluate the Given limit:

At $x=2$, the value of the given rational function takes the form $\frac{0}{0}$.

$\therefore \lim _{x \rightarrow 3} \frac{x^{4}-81}{2 x^{2}-5 x-3}=\lim _{x \rightarrow 3} \frac{(x-3)(x+3)\left(x^{2}+9\right)}{(x-3)(2 x+1)}$

$=\lim _{x \rightarrow 3} \frac{(x+3)\left(x^{2}+9\right)}{2 x+1}$

$=\frac{(3+3)\left(3^{2}+9\right)}{2(3)+1}$

$=\frac{6 \times 18}{7}$

$=\frac{108}{7}$