Question

$\large\boxed{ \rm~limites~}$

$\large \boxed{ \begin{array}{l} \rm\dfrac{lim}{x\rightarrow8} ~ \dfrac{x {}^{2} - 64}{x - 8} \end{array}}$
ᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠ​

Answer 1

Explicação passo-a-passo:

$\large \boxed{ \begin{array}{l} \rm\dfrac{lim}{x\rightarrow8} ~ \dfrac{x {}^{2} - 64}{x - 8}=\dfrac{(x-8).(x+8)}{(x-8)}=(x+8)\end{array}}$

$\large \boxed{ \begin{array}{l} \rm\dfrac{lim}{x\rightarrow8} ~ \dfrac{x {}^{2} - 64}{x - 8}=8+8=16 \end{array}}$

ᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠᅠ

Answer 2

$\large \boxed{ \begin{array}{l} \rm\dfrac{lim}{x\rightarrow8} ~ \dfrac{x {}^{2} - 64}{x - 8} \\ \\ \rm\dfrac{lim}{x\rightarrow8} \: \: \dfrac{x {}^{2} - 64 }{x - 8} \\ \\ \rm \dfrac{lim}{x\rightarrow8} \: \: x - 8 \\ \\ 0 \\ \\ 0 \\ \\ \rm \dfrac{lim}{x\rightarrow8 } \: \: \dfrac{(x - 8) \times (x + 8)}{x - 8} \\ \\ \rm \dfrac{lim}{x\rightarrow8} \: \: x + 8 \\ \\ 8 + 8 \\ \\ \boxed{ \boxed{16}} \\ \\ \\ \\ \large \boxed{ \begin{array}{l} \rm\dfrac{lim}{x\rightarrow8} ~ \dfrac{x {}^{2} - 64}{x - 8} = \boxed{ \boxed{ \boxed{16}}}\end{array}} \end{array}}$