Question

Tentukan nilai dari limit berikut!
[tex]\displaystyle\lim_{x \to \infty} (x + 2) - \sqrt{{x}^{2} + x - 2}[/tex]
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Selamat Berjuang​

Answer 1

Verified answer

Jawab:

[tex]= \lim_{x \to \infty} (x+2)-\sqrt{x^2+x-2} \\\\= \lim_{x \to \infty} (x+2)-\sqrt{x^2+x-2}\times \frac{(x+2)+\sqrt{x^2+x-2} \\}{(x+2)+\sqrt{x^2+x-2} \\} \\\\=\frac{(x^2+4x+4)-(x^2+x-2)}{ (x+2)+\sqrt{x^2+x-2}}\\\\ = \frac{3x+2}{x+2+\sqrt{x^2+x-2}}\\\\=\frac{3x}{x+\sqrt{x^2} } \\\\=\frac{3x}{x+x} \\\\=\frac{3x}{2x}\\\\=\frac{3}{2}[/tex]

Answer 2

∞ Limit Tak Hingga ∞

[tex]\sf \lim\limits_{x \to \infty} ((x+2) -\sqrt{x^{2} +x-2} )\\\\=\lim\limits_{x \to \infty} (\sqrt{(x+2)^{2} } -\sqrt{x^{2} +x-2} )\\\\=\lim\limits_{x \to \infty} (\sqrt{x^{2} +4x+4 } -\sqrt{x^{2} +x-2} )[/tex]

Jika bentuk  [tex]\sf \lim\limits_{x \to \infty} (\sqrt{ax^{2}+bx+c } -\sqrt{px^{2} +qx+r} )[/tex]  dengan a = p

Maka hasilnya [tex]\sf \frac{b-q}{2\sqrt{a} }[/tex] .

[tex]\sf\\=\frac{4-1}{2\sqrt{1} }\\\\=\frac{3}{2\cdot 1} \\\\\boxed{\sf =\frac{3}{2} }[/tex]

Jadi, hasil limit tak hingga tersebut adalah 3/2.

[tex]\boxed{\sf{shf}}[/tex]