Question

Evaluate the limit : $\underset{x\to 4}{lim}\frac{x^2-7x+12}{x^2-3x-4}$

Answer 1

Given : $\underset{x\to 4}{lim}\frac{x^2-7x+12}{x^2-3x-4}becomes\frac{0}{0}form$

Using Factorization method

$\Rightarrow \underset{x\to 4}{lim}\frac{x^2-7x+12}{x^2-3x-4}$

$\Rightarrow \underset{x\to 4}{lim}\frac{x^2-4x-3x+12}{x^2-4x+x-4}$

$\Rightarrow \underset{x\to 4}{lim}\frac{x(x-4)-3(x-4)}{x(x-4)+(x-4)}$

$\Rightarrow \underset{x\to 4}{lim}\frac{(x-3)(x-4)}{(x-4)(x+1)}\{\because x\ne 4\}$

$\Rightarrow \underset{x\to 4}{lim}\frac{(x-3)}{(x+1)}$

$\Rightarrow \frac{4-3}{4+1}=\frac{1}{5}$

Therefore,

$\underset{x\to 4}{lim}\frac{x^2-7x+12}{x^2-3x-4}=\frac{1}{5}$