Question

Evaluate the limit: $\underset{x\to 2}{lim}\frac{x^3-8}{x^2-4}$

Answer 1

Given: $\underset{x\to 2}{lim}\frac{x^3-8}{x^2-4}becomes\frac{0}{0}form$

Using Factorization method

$\Rightarrow \underset{x\to 2}{lim}\frac{x^3-8}{x^2-4}=\underset{x\to 2}{lim}\frac{x^3-2^3}{x^2-2^2}$

$[\because (a^3-b^3)=(a-b\left)\right(a^2+ab+b^2\left)\right]$

$\Rightarrow \underset{x\to 2}{lim}\frac{(x-2)(x^2+2x+2^2)}{(x+2)(x-2)}$

$\Rightarrow \underset{x\to 2}{lim}\frac{(x^2+2x+4)}{(x+2)}$ $[x\ne 2]$

$=\frac{2^2+2\left(2\right)+4}{2+2}$

$=\frac{12}{4}=3$

Therefore,

$\Rightarrow \underset{x\to 2}{lim}\frac{x^3-8}{x^2-4}=3$