LiMIt Bentuk ~ - ~
Penjelasan dengan langkah-langkah:
[tex]\sf lim_{x\to \infty}\sqrt{x^2+2x} - \sqrt {x^2 - x}[/tex]
kalikan akar sekawan, maka
[tex]\sf lim_{x\to \infty}\ \dfrac{(\sqrt{x^2+2x} - \sqrt {x^2 - x})(\sqrt{x^2+2x} + \sqrt {x^2 - x})}{(\sqrt{x^2+2x} +\sqrt {x^2 - x})}[/tex]
[tex]\sf lim_{x\to \infty}\ \dfrac{x^2+2x- x^2 +x}{(\sqrt{x^2+2x} +\sqrt {x^2 - x})}[/tex]
[tex]\sf lim_{x\to \infty}\ \dfrac{3x}{(\sqrt{x^2+2x} +\sqrt {x^2 - x})}[/tex]
[tex]\sf lim_{x\to \infty}\ \dfrac{\sqrt{(3x)^2}}{(\sqrt{x^2+2x} +\sqrt {x^2 - x})}[/tex]
[tex]\sf lim_{x\to \infty}\ \dfrac{\sqrt{9x^2}}{(\sqrt{x^2+2x} +\sqrt {x^2 - x})}[/tex]
x tertinggi pangkat 2
[tex]\sf lim_{x\to \infty}\ \dfrac{\sqrt{9x^2}}{(\sqrt{x^2} +\sqrt {x^2 })}= \dfrac{\sqrt 9}{\sqrt 1 +\sqrt 1} = \dfrac{3}{1+1}= \dfrac{3}{2}[/tex]
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cara rumus
[tex]\lim_{x\to \infty}\sqrt{ax^2+ bx+ c} - \sqrt{px^2 +qx + r}},[/tex]
jika a= p, maka rumus limit =
[tex]\sf limit =\dfrac{b-q}{2\sqrt a}[/tex]
soal
[tex]\sf lim_{x\to \infty}\sqrt{x^2+2x} - \sqrt {x^2 - x}[/tex]
a= 1, b = 2 , p= 1 , q = -1
[tex]\sf limit =\dfrac{b-q}{2\sqrt a} = \dfrac{2-(-1)}{2\sqrt 1} =\dfrac{3}{2}[/tex]
[answer.2.content]
