Matematika

Pertanyaan

Limit aljabar fungsi tak hingga
Limit aljabar fungsi tak hingga

1 Jawaban

  • [tex]L= \frac{12.6^{x}. \frac{1}{6}+36.3^{x}. \frac{1}{3} +2017 }{6.3^{x}.3-15.6^{x}. \frac{1}{6}-2015 } . \frac{6}{6} [/tex]

    [tex]L= \frac{12.6^{x}+72.3^{x}.+2017.6 }{108.3^{x}-15.6^{x}-2015.6 } [/tex]

    [tex]L= \frac{12. \frac{6^{x}}{3^{x}} +72. \frac{3^{x}}{3^{x}} + \frac{2017.6}{3^{x}} }{108. \frac{3^{x}}{3^{x}} -15. \frac{6^{x}}{3^{x}} - \frac{2015.6}{3^{x}} } [/tex]

    [tex]L= \frac{12. 2^{x} +72. + \frac{2017.6}{3^{x}} }{108 -15. 2^{x} - \frac{2015.6}{3^{x}} } [/tex]

    [tex]L= \frac{12. \frac{2^{x} }{2^{x} } + \frac{72}{2^{x} } + \frac{2017.6}{6^{x}} }{ \frac{108}{2^{x} } -15. \frac{2^{x}}{2^{x}} - \frac{2015.6}{6^{x}} } [/tex]

    [tex]L= \frac{12. + \frac{72}{2^{x} } + \frac{2017.6}{6^{x}} }{ \frac{108}{2^{x} } -15. - \frac{2015.6}{6^{x}} } [/tex]

    [tex]L= \frac{12. + 0 + 0 }{0 -15. -0} [/tex]

    [tex]L=- \frac{4}{5} [/tex]