Question

Alguem Me da um hellp :D Lim ( (f(x +h) - f(x) ) /0 ) Sendo f(x) = -x² -x h-0

Answer 1

Resposta:

resposta  -2x - 1

Explicação passo-a-passo:

f(x) = - x² - x

lim  [ f(x+h) - f(x) ] / h

h -> 0

lim [ ( -1*(x+h)² -1 *(x+h) - ( -x² - x) ] / h

h -> 0

lim [ -x² -2xh -h² ) -x -h + x² + x ] / h

h -> 0

lim [ -x² + x² -2xh -h² -h  -x  + x ] / h

h -> 0

lim [ -2xh -h² -h  ] / h

h -> 0

lim  h/h [ -2x -h -1 ]

h -> 0

lim  [ -2x -1 ]

h -> 0

resposta -2x - 1

Answer 2

Explicação passo-a-passo:

$\sf \lim_{h\to~0}=\dfrac{f(x+h)-f(x)}{h}$

$\sf \lim_{h\to~0}=\dfrac{-(x+h)^2-(x+h)-(-x^2-x)}{h}$

$\sf \lim_{h\to~0}=\dfrac{-(x^2+2xh+h^2)-x-h+x^2+x}{h}$

$\sf \lim_{h\to~0}=\dfrac{-x^2-2xh-h^2-x-h+x^2+x}{h}$

$\sf \lim_{h\to~0}=\dfrac{-x^2+x^2-2xh-h-h^2-x+x}{h}$

$\sf \lim_{h\to~0}=\dfrac{-2xh-h-h^2}{h}$

$\sf \lim_{h\to~0}=\dfrac{h\cdot(-2x-1-h)}{h}$

$\sf \lim_{h\to~0}=-2x-1-h$

$\sf =-2x-1-0$

$\sf =\red{-2x-1}$