Matematika Sekolah Menengah Atas 1. Nilai dari lim x→♾️ (√3x+1 - √3x+2)=

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1. Nilai dari lim x→♾️ (√3x+1 - √3x+2)=

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[tex] = \lim \limits_{x \to \infty }( \sqrt{3x + 1} - \sqrt{3x + 2} )[/tex]

[tex]= \lim \limits_{x \to \infty }( \sqrt{3x + 1} - \sqrt{3x + 2} ) \times \frac{ \sqrt{3x + 1} + \sqrt{3x + 2} }{ \sqrt{3x + 1} + \sqrt{3x + 2} } \\ [/tex]

[tex]= \lim \limits_{x \to \infty } \frac{( \sqrt{3x + 1}) {}^{2} - (\sqrt{3x + 2}) {}^{2} }{ \sqrt{3x + 1} + \sqrt{3x + 2} } \\ [/tex]

[tex]= \lim \limits_{x \to \infty } \frac{({3x + 1}) - ({3x + 2}) }{ \sqrt{3x + 1} + \sqrt{3x + 2} } \\ [/tex]

[tex]= \lim \limits_{x \to \infty } \frac{ - 1 }{ \sqrt{3x + 1} + \sqrt{3x + 2} } \\ [/tex]

[tex] = \lim \limits_{x \to \infty } \frac{ - \frac{1}{ \sqrt{x} } }{ \sqrt{3 + \frac{1}{x} } + \sqrt{3+ \frac{2}{x} } } \times \frac{ \sqrt{x} }{ \sqrt{x} } [/tex]

[tex] = \lim \limits_{x \to \infty } \frac{ - \frac{1}{ \sqrt{x} } }{ \sqrt{3 + \frac{1}{x} } + \sqrt{3+ \frac{2}{x} } } [/tex]

[tex] = \frac{ - \frac{1}{ \sqrt{ \infty } } }{ \sqrt{3 + \frac{1}{ \infty } } + \sqrt{3 + \frac{2}{ \infty } } } [/tex]

[tex] = \frac{0}{ \sqrt{3} + \sqrt{3} } [/tex]

[tex] = 0[/tex]

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