Question

Solve
$\underset{x\to 3}{lim}\frac{x^3-7x^2+15x-9}{x^4-5x^3+27x-27}$

Answer 1

$\underset{x\to 3}{lim}\frac{x^3-7x^2+15x-9}{x^4-5x^3+27x-27}\left(\frac{0}{0}\right)$ Using $L$Hospital rule

$\underset{x\to 3}{lim}\frac{3x^2-14x+15}{4x^3-15x^2+27}$

$\underset{x\to 3}{lim}\frac{6x-14}{{12}^2-30x}$

$\frac{6\times 1-14}{12\times 9-30\times 3}=\frac{4}{16}=\frac{2}{9}$