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# Evaluate the following limits  $\lim\limits_{x \to 3} \large\frac{x^4-81}{2x^2-5x-3}$

$\begin{array}{1 1}\large\frac{108}{7} \\ 0 \\ \infty \\ \large\frac{18}{7} \end{array}$

At $x = 2$ the rational function takes the form $\large\frac{0}{0}$
$\Rightarrow \lim\limits_{x \to 3} \large\frac{x^4-81}{2x^2-5x-3}$ $= \lim\limits_{x \to 3} \large\frac{(x-3)(x+3)(x^2+9)}{(x-3)(2x+1)}$
$\qquad = \lim\limits_{x \to 3} \large\frac{(x+3)(x^2+9)}{2x+1}$
$\qquad = \large\frac{(3+3)(3^2+9)}{2(3)+1}$
$\qquad = \large\frac{6 \times 18}{7}$ $= \large\frac{108}{7}$
edited May 16, 2014