Question

Evaluate the limit : $\underset{x\to 3}{lim}\frac{x^4-81}{x^2-9}$

Answer 1

Given : $\underset{x\to 3}{lim}\frac{x^4-81}{x^2-9}becomes\frac{0}{0}form$

Using Factorization method

$\Rightarrow \underset{x\to 3}{lim}\frac{x^4-81}{x^2-9}=\underset{x\to 3}{lim}\frac{(x^2{)}^2-9^2}{x^2-9}$

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

$[\because (a^2-b^2)=(a+b\left)\right(a-b\left)and\right\{x^2\ne 9\left\}\right]$

$\Rightarrow \underset{x\to 3}{lim}(x^2+9)$

$=(3^2+9)=18$

Therefore,

$\Rightarrow \underset{x\to 3}{lim}\frac{x^4-81}{x^2-9}=18$