Question

Evaluate: ${lim}_{x\to 3}\frac{x^4-81}{2x^2-5x-3}$

Answer 1

${lim}_{x\to 3}\frac{x^4-81}{2x^2-5x-3}$

$={lim}_{x\to 3}\frac{(x^2-9)(x^2+9)}{2x^2-6x+x-3}$

$={lim}_{x\to 3}\frac{(x-3)(x+3)(x^2+9)}{2x(x-3)+1(x-3)}$

$={lim}_{x\to 3}\frac{(x-3)(x+3)(x^2+9)}{(x-3)(2x+1)}$

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

$=\frac{6\times 18}{7}=\frac{108}{7}$