Question

Evaluate the given limit :
$\underset{x\to 3}{lim}\frac{x^4-81}{2x^2-5x-3}$

Answer 1

Given $\underset{x\to 3}{lim}\frac{x^4-81}{2x^2-5x-3}$
Substituting $x=3$
$=\frac{(3{)}^4-81}{2(3{)}^2-5(3)-3}$
$=\frac{81-81}{18-15-3}=\frac{0}{0}$
Since it is a $\frac{0}{0}$ form

We simplify as
$\underset{x\to 3}{lim}\frac{x^4-81}{2x^2-5x-3}$= $\underset{x\to 3}{lim}\frac{(x^2{)}^2-(9{)}^2}{2x^2-6x+x-3}$

Using $a^2-b^2=(a-b)(a+b)$
=$\underset{x\to 3}{lim}\frac{(x^2-9)(x^2+9)}{2x(x-3)+1(x-3)}$
=$\underset{x\to 3}{lim}\frac{(x^2-(3{)}^2\left)\right(x^2+9)}{(2x+1)(x-3)}$

Using $a^2-b^2=(a-b)(a+b)$
=$\underset{x\to 3}{lim}\frac{(x-3)(x+3)(x^2+9)}{(2x+1)(x-3)}$
=$\underset{x\to 3}{lim}\frac{(x+3)(x^2+9)}{2x+1}(\because x\ne 3)$
=$\underset{x\to 3}{lim}\frac{(x+3)(x^2+9)}{2x+1}$
Substituting $x=3$
$=\frac{(3+3)\left(\right(3{)}^2+9)}{2\times 3+1}$
$=\frac{6(9+9)}{6+1}$
$=\frac{6\left(18\right)}{7}$
$=\frac{108}{7}$
$\therefore$ $\underset{x\to 3}{lim}\frac{x^4-81}{2x^2-5x-3}=\frac{108}{7}$