Ensino Superior ⇒ Limites

[jose carlos de almeida]

Calcule o limite:

• [tex3]\lim_{x\to 0}\frac{\sqrt[3]{8+h}-2}{h}[/tex3]

  Código: Selecionar tudoSHIFT+Click na eq = ZOOM

  [tex3]\lim_{x\to 0}\frac{\sqrt[3]{8+h}-2}{h}[/tex3]

[Alexandre_SC]

• [tex3]\lim_{x\to 0}\ \frac{\sqrt[3]{8+h}-2}{h} =\lim_{x\to 0}\frac{\sqrt[3]{8+h}-2}{h}\cdot \frac{ \sqrt[3]{(8+h)^2} +2\sqrt[3]{8+h}+4}{\sqrt[3]{(8+h)^2} +2\sqrt[3]{8+h}+4}=[/tex3]

  Código: Selecionar tudoSHIFT+Click na eq = ZOOM

  [tex3]\lim_{x\to 0}\ \frac{\sqrt[3]{8+h}-2}{h} =\lim_{x\to 0}\frac{\sqrt[3]{8+h}-2}{h}\cdot \frac{ \sqrt[3]{(8+h)^2} +2\sqrt[3]{8+h}+4}{\sqrt[3]{(8+h)^2} +2\sqrt[3]{8+h}+4}=[/tex3]

• [tex3]\lim_{x\to 0}\ \frac{8+h-8}{h\cdot( \sqrt[3]{(8+h)^2} +2\sqrt[3]{8+h}+4)}=\lim_{x\to 0} \ \frac{1}{\sqrt[3]{(8+h)^2} +2\sqrt[3]{8+h}+4}=\frac{1}{12}.[/tex3]

  Código: Selecionar tudoSHIFT+Click na eq = ZOOM

  [tex3]\lim_{x\to 0}\ \frac{8+h-8}{h\cdot( \sqrt[3]{(8+h)^2} +2\sqrt[3]{8+h}+4)}=\lim_{x\to 0} \ \frac{1}{\sqrt[3]{(8+h)^2} +2\sqrt[3]{8+h}+4}=\frac{1}{12}.[/tex3]