Matematika

Pertanyaan

lim x memdekati 4
x^2 - 16 / akar(32 - x^2) - akar(2x + 8)

1 Jawaban

  • [tex]
    \lim_{x\to4}\frac{x^2-16}{\sqrt{32-x^2}-\sqrt{2x+8}}\cdot\frac{\sqrt{32-x^2}+\sqrt{2x+8}}{\sqrt{32-x^2}+\sqrt{2x+8}}\\
    =\lim_{x\to4}\frac{(x^2-16)\left(\sqrt{32-x^2}+\sqrt{2x+8}\right)}{(32-x^2)-(2x+8)}\\
    =\lim_{x\to4}\frac{(x+4)(x-4)\left(\sqrt{32-x^2}+\sqrt{2x+8}\right)}{-(x+6)(x-4)}\\
    =\lim_{x\to4}\frac{(x+4)\left(\sqrt{32-x^2}+\sqrt{2x+8}\right)}{-(x+6)}\\
    =\frac{(4+4)\left(\sqrt{32-4^2}+\sqrt{2(4)+8}\right)}{-(4+6)}\\
    =\frac{8\cdot(4+4)}{-10}\\
    =-6,4[/tex]

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