Question

If $l=\underset{x\to 3}{\lim }\frac{x^3-7x^2+15x-9}{x^4-5x^3+27x-27}$, then find the value of $36l-7$.

Answer 1

Answer: 1

$l=\underset{x\to 3}{\lim }\frac{x^3-7x^2+15x-9}{x^4-5x^3+27x-27}$
$=\underset{x\to 3}{\lim }\frac{(x-3)\left(x^2-4x+3\right)}{(x-3)\left(x^3-2x^2-6x+9\right)}$
$=\underset{x\to 3}{\lim }\frac{x^2-4x+3}{x^3-2x^2-6x+9}$
$=\underset{x\to 3}{\lim }\frac{(x-1)(x-3)}{(x-3)\left(x^2+x-3\right)}$
$=\underset{x\to 3}{\lim }\frac{x-1}{x^2+x-3}=\frac{2}{9}$
$\Rightarrow 36l-7=36\left(\frac{2}{9}\right)-7$
$=8-7=1$