Question

Evaluate $\underset{x\to 3}{lim}\left(\frac{x^4-81}{2x^2-5x-3}\right)$

Answer 1

$\underset{x\to 3}{lim}\left(\frac{x^4-81}{2x^2-5x-3}\right)$

The expression takes the form $\frac{0}{0}$

Using Lhospital~s Rule

$\underset{x\to 3}{lim}\frac{4x^3}{4x-5}$
$\frac{4(3{)}^3}{4\left(3\right)-5}$
$\frac{4\times 27}{12-5}=\frac{108}{7}$