[tex] (\sqrt{{x}^{2} } - 6x + 5 \: - \sqrt{ {x}^{2} } + x - 1)[/tex]
bantu jawab..
•kelas IX Mipa
•Matematika wajib
•Bab Limit fungsi
•Limit tak hingga dalam bentuk akar
Penjelasan dengan langkah-langkah:
[tex] = \lim \limits_{x \to \infty } ( \sqrt{ {x}^{2} - 6x + 5 } - \sqrt{ x^2 + x - 1} )[/tex]
[tex] = \frac{\sqrt{x^2 - 6x + 5}^2 - \sqrt{x^2+ x - 1}^2}{ \sqrt{ {x}^{2} - 6x + 5} + \sqrt{ {x}^{2} + x - 1} } [/tex]
[tex] = \frac{( {x}^{2} - 6x + 5) - ( {x}^{2} + x - 1 ) }{ \sqrt{ {x}^{2} - 6x + 5 } + \sqrt{ {x}^{2} + x - 1} } [/tex]
[tex] = \frac{ - 7x + 6}{ \sqrt{ {x}^{2} - 6x + 5} + \sqrt{ {x}^{2} + x - 1 } } [/tex]
[tex] = \frac{ - 7 + \frac{6}{x} }{ \sqrt{1 - \frac{6}{x} + \frac{5}{ {x}^{2} } } + \sqrt{1 + \frac{1}{x} - \frac{1}{ {x}^{2} } } } [/tex]
[tex] = \lim \limits_{x \to \infty } \frac{ - 7 + \frac{6}{x} }{ \sqrt{1 - \frac{6}{x} + \frac{5}{ {x}^{2} } } + \sqrt{1 + \frac{1}{x} - \frac{1}{ {x}^{2} } } } [/tex]
[tex] = \frac{ - 7 + 0}{ \sqrt{1 - 0 + 0} + \sqrt{1 + 0 - 0} } [/tex]
[tex] = \frac{ - 7}{1 + 1} [/tex]
[tex] = - \frac{7}{2} [/tex]
[answer.2.content]
