Tolongnya yang jawab saya jadikan jawaban terbaik .harus benar no: 5 dan 6.... Question from @Qorysweet

December 2018 1 20 Report

Tolongnya yang jawab saya jadikan jawaban terbaik .harus benar no: 5 dan 6

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Nomor 5

$\displaystyle ~~~\lim_{x\to\,\infty}(\sqrt{x+\sqrt{2x}}-\sqrt{x})\times\frac{(\sqrt{x+\sqrt{2x}}+\sqrt{x})}{(\sqrt{x+\sqrt{2x}}+\sqrt{x})}\\\\=\lim_{x\to\,\infty}\frac{x+\sqrt{2x}-x}{(\sqrt{x+\sqrt{2x}}+\sqrt{x})}\\\\=\lim_{x\to\,\infty}\frac{\sqrt{2x}}{(\sqrt{x+\sqrt{2x}}+\sqrt{x})}\\\\\\=\lim_{x\to\,\infty}\frac{\sqrt{\dfrac{2x}{x}}}{(\sqrt{\dfrac{x}{x}+\sqrt{\dfrac{2x}{x^2}}}+\sqrt{\dfrac{x}{x}})}\\\\\\=\frac{\sqrt{2}}{\sqrt{1+0}+1}\\\\=\boxed{\frac{1}{2}\sqrt{2}}$
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Nomor 6

Soal bisa kita ubah menjadi:

$\displaystyle \lim_{x\to ~}(\sqrt{x^2-2x+6}-x)+(\sqrt{x^2+2x-3}-x)-(\sqrt{4x^2+16x-8}-2x)\\~~~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~~~~~~~~~~~~~\downarrow~~~~~~~~~~~~~~~~~~~~~~~~~~~\downarrow\\~~~~~~~~~~~~~~~~\boxed{A}~~~~~~~~~~~~~~~~~~~~~~~\boxed{B}~~~~~~~~~~~~~~~~~~~~~~~~~~\boxed{C}$

$\displaystyle A=\lim_{x\to\,\infty}(\sqrt{x^2-2x+6}-x)\times\frac{\sqrt{x^2-2x+6}+x}{\sqrt{x^2-2x+6}+x}\\\\~~~=\lim_{x\to\,\infty}\frac{x^2-2x+6-x^2}{\sqrt{x^2-2x+6}+x}\\\\~~~=\lim_{x\to\,\infty}\frac{\boxed{-2}x+6}{\sqrt{\boxed{1}~x^2-2x+6}+\boxed{1}~x}\\\\\\~~~=\frac{-2}{1+1}\\\\A=-1$

$\displaystyle B=\lim_{x\to\,\infty}(\sqrt{x^2+2x-3}-x)\times\frac{\sqrt{x^2+2x-3}+x}{\sqrt{x^2+2x-3}+x}\\\\~~~=\lim_{x\to\,\infty}\frac{x^2+2x-3-x^2}{\sqrt{x^2+2x-3}+x}\\\\~~~=\lim_{x\to\,\infty}\frac{\boxed{2}x-3}{\sqrt{\boxed{1}~x^2+2x-3}+\boxed{1}~x}\\\\\\~~~=\frac{2}{1+1}\\\\B=1$

$\displaystyle C=\lim_{x\to\,\infty}(\sqrt{4x^2+16x-8}-2x)\times\frac{\sqrt{4x^2+16x-8}+2x}{\sqrt{4x^2+16x-8}+2x}\\\\~~~=\lim_{x\to\,\infty}\frac{4x^2+16x-8-4x^2}{\sqrt{4x^2+16x-8}+2x}\\\\~~~=\lim_{x\to\,\infty}\frac{\boxed{16}x-8}{\sqrt{\boxed{4}~x^2+16x-3}+\boxed{2}~x}\\\\\\~~~=\frac{16}{2+2}\\\\C=4$

Jadi Hasil akhir
A + B - C = -1 + 1 - 4 = -4 (Jawab:E)

2 votes Thanks 2

Tolong no 1 aja, besok dikumpul, tapi harus benar ya

Mohon bantuannyano 3 . terimakasih

Mohon bantuannya no 1 dan 2. cepatnya

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