Question

Calcule o limite de deltaX ->0
( f(x + deltaX ) - f(x) / deltaX)
para f(x) = 2* raiz quadrada de x

Answer 1

Explicação passo-a-passo:

$\sf \lim_{\Delta x~\Rightarrow~0}~\dfrac{f(x+\Delta x)-f(x)}{\Delta x}$

$\sf \lim_{\Delta x~\Rightarrow~0}~\dfrac{2\cdot\sqrt{x+\Delta x}-2\sqrt{x}}{\Delta x}$

$\sf \lim_{\Delta x~\Rightarrow~0}~\dfrac{2\cdot(\sqrt{x+\Delta x}-\sqrt{x})}{\Delta x}$

$\sf \lim_{\Delta x~\Rightarrow~0}~\dfrac{2\cdot(\sqrt{x+\Delta x}-\sqrt{x})}{\Delta x}\cdot\dfrac{(\sqrt{x+\Delta x}+\sqrt{x})}{(\sqrt{x+\Delta x}+\sqrt{x})}$

$\sf \lim_{\Delta x~\Rightarrow~0}~\dfrac{2\cdot[(\sqrt{x+\Delta x})^2-(\sqrt{x})^2]}{\Delta x}$

$\sf \lim_{\Delta x~\Rightarrow~0}~\dfrac{2\cdot(x+\Delta x-x)}{\Delta x\cdot(\sqrt{x+\Delta x}+\sqrt{x})}$

$\sf \lim_{\Delta x~\Rightarrow~0}~\dfrac{2\cdot\Delta x}{\Delta x\cdot(\sqrt{x+\Delta x}+\sqrt{x})}$

$\sf \lim_{\Delta x~\Rightarrow~0}~\dfrac{2}{\sqrt{x+\Delta x}+\sqrt{x}}$

$\sf =\dfrac{2}{\sqrt{x+0}+\sqrt{x}}$

$\sf =\dfrac{2}{\sqrt{x}+\sqrt{x}}$

$\sf =\dfrac{2}{2\sqrt{x}}$

$\sf =\red{\dfrac{1}{\sqrt{x}}}$