Question

$\lim_{x \to \ 8}$ $\frac{\sqrt{2+\sqrt[3]{x} } - 2 }{(x - 8)^{2} }$

Answer 1

Resposta:

lim    [√(2+∛x) -2]/(x-8)²

x-->8

lim    [√(2+∛x) -2]*[√(2+∛x) +2]/(x-8)²[√(2+∛x) +2]

x-->8

lim    [√(2+∛x)² -2²]/(x-8)²[√(2+∛x) +2]

x-->8

lim    [2+∛x -4]/(x-8)²[√(2+∛x) +2]

x-->8

lim    [∛x -2]/(x-8)²[√(2+∛x) +2]

x-->8

***[∛x -2]³=x-6∛x²+12∛x-8

***[∛x -2]³=x-8 -6∛x²+12∛x

***[∛x -2]³=x-8 -6∛x*(∛x-2)

***x-8=[∛x -2]³+6∛x*(∛x-2)

***x-8=[∛x -2]*[(∛x -2)²+6∛x]

***∛x -2=(x-8)/[(∛x -2)²+6∛x]

lim    {(x-8)/[(∛x -2)²+6∛x]} /(x-8)²[√(2+∛x) +2]

x-->8⁺

lim    {1/[(∛x -2)²+6∛x]} /(x-8)[√(2+∛x) +2]

x-->8⁺

= {1/[(∛8 -2)²+6∛8]} /(8-8)[√(2+∛8) +2]

= {1/[(∛8 -2)²+6∛8]} /(8-8)[√(2+∛8) +2]

= {1/[0+6∛8]} /(8-8)[2 +2]

= {1/[0+6*2]} /(8-8)[4]

= 1/[0+6*2]} /(0⁺)[4]

= 1/0⁺

= +∞

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lim    {(x-8)/[(∛x -2)²+6∛x]} /(x-8)²[√(2+∛x) +2]

x-->8⁻

lim    {1/[(∛x -2)²+6∛x]} /(x-8)[√(2+∛x) +2]

x-->8⁻

= {1/[(∛8 -2)²+6∛8]} /(8-8)[√(2+∛8) +2]

= {1/[0+6∛8]} /(8-8)[2 +2]

= {1/[0+6*2]} /(8-8)[4]

= 1/[0+6*2]} /(0⁻)[4]

= 1/0⁻

=  -∞

Os limites Laterais são diferentes, este limite não existe