Answer
$\lim\limits_{x\to3}{\frac{x^4-81}{x-3}} =108$
Work Step by Step
$\lim\limits_{x\to3}{\frac{x^4-81}{(x-3)}} = \lim\limits_{x\to 3}{\frac{(x^2-9)(x^2+9)}{(x-3)}} =\lim\limits_{x\to 3}{\frac{(x-3)(x+3)(x^2+9)}{(x-3)}} =\lim\limits_{x\to 3}{\frac{(x+3)(x^2+9)}{1}}=(\lim\limits_{x\to 3}x+3)(\lim\limits_{x\to 3}x^2 +9)=6\times18=108$
