Question

resolva o limite
$\lim_{x \to \ 1} \frac{x^{2} -4x+3}{x-1}$

Answer 1

Explicação passo-a-passo:

$lim_{x - > 1}( \frac{ {x}^{2} - 4x + 3}{x - 1} )$

$lim_{x - > 1}( \frac{ {x}^{2} - 3x - x + ( - 1)( - 3) }{ x - 1} )$

$lim_{x - > 1}( \frac{x(x - 3) - x + ( - 1)( - 3)}{x - 1} )$

$lim_{x - > 1}( \frac{x(x - 3) -( - 1)(x - 3)}{x - 1} )$

$lim_{x - > 1}( \frac{(x - 1)(x - 3)}{x - 1} )$

$lim_{x - > 1}( x - 3)$

$= 1 - 3$

$= - 2$

Answer 2

ax²+bx+c=a(x-x')(x-x")

${x}^{2} - 4x + 3 = 0 \\ \Delta = {b}^{2} - 4ac \\ \Delta = {( - 4)}^{2} - 4.1.3 \\ \Delta = 16 - 12 \\ \Delta = 4$

$x = \frac{-b±\sqrt{\Delta}}{2a} \\ x = \frac{ - ( - 4)± \sqrt{4}}{2.1} \\ x = \frac{4±2}{2}$

$x' = \frac{4 + 2}{2} = \frac{6}{2} = 3 \\ x'' = \frac{4 - 2}{2} = \frac{2}{2} = 1$

${x}^{2} - 4x + 3 = (x - 1)(x - 3)$

$\lim_{x \to \ 1} \frac{x^{2} -4x+3}{x-1} \\ \lim_{x \to \ 1} \frac{(x - 1)(x - 3)}{x-1} \\ \lim_{x \to \ 1}x - 3 = 1 - 3 = - 2$