Question

o valor de lim x-->3 $\frac{x^{2}+x-12 }{x-3}$ e:

Answer 1

Explicação passo-a-passo:

$\lim_{x\to~3}~\dfrac{x^2+x-12}{x-3}=\lim_{x\to~3}~\dfrac{(x-3)\cdot(x+4)}{x-3}$

$\lim_{x\to~3}~\dfrac{x^2+x-12}{x-3}=\lim_{x\to~3}~(x+4)$

$\lim_{x\to~3}~\dfrac{x^2+x-12}{x-3}=3+4$

$\lim_{x\to~3}~\dfrac{x^2+x-12}{x-3}=7$

Answer 2

Explicação passo-a-passo:

Olá!!!

$lim \: \frac{ {x}^{2} + x - 12}{x - 3} \\ x = > 3 \\ \\ lim \: \frac{( {3}^{2} + 3 - 12) }{(3 - 3)} \\ \\ lim \: \frac{( 9+ 3 - 12) }{(0)} \\ \\ lim \: \frac{( 12 - 12) }{(3 - 3)} \\ \\ lim \: \frac{( 0) }{(0)} \\ \\ indeterminado \\ \\ lim \: \frac{ {x}^{2} + 4x - 3x - 12}{x - 3} \\ x = > 3 \\ \\lim \: \frac{ x (x + 4)- 3(x + 4)}{x - 3} \\ x = > 3 \\ \\ lim \: \frac{ \cancel{ (x - 3)}(x + 4)}{ \cancel{(x - 3)}} \\ x = > 3 \\ \\ lim \: (x + 4) \\ x = > 3 \\ \\ lim \: (7) \\ x = > 3 \\ \\ \checkmark \boxed{ \boxed{lim \rightarrow \boxed{7}}}$