(³log 27 + sin² x + cos² x) Lim x -> 4. X² - 6x + 11 = tolong bantu jawab, terimakasih
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[tex]\begin{aligned}&\left({}^3\log27+\sin^2x+\cos^2x\right)\left(\lim_{x\to4}x^2-6x+11\right)\\&=\boxed{\ \bf12\ }\end{aligned}[/tex]
Pembahasan
[tex]\begin{aligned}&\left({}^3\log27+\sin^2x+\cos^2x\right)\left(\lim_{x\to4}x^2-6x+11\right)\\&\quad\left[\ \begin{aligned}\bullet\ &{}^3\log27={}^3\log3^3=3\\\bullet\ &\sin^2x+\cos^2x=1\end{aligned}\right.\\&{=\ }(3+1)\left(\lim_{x\to4}x^2-6x+11\right)\\&{=\ }(4)\left(\lim_{x\to4}x^2-6x+11\right)\end{aligned}[/tex]
[tex]\begin{aligned}&\quad\left[\ \begin{aligned}&\textsf{limit tersebut bukan bentuk tak tentu}\\&\textsf{sehingga bisa langsung substitusi $x$}\end{aligned}\right.\\&{=\ }(4)\left(4^2-6(4)+11\right)\\&{=\ }(4)(16-24+11)\\&{=\ }(4)(3)\\&{=\ }\boxed{\ \bf12\ }\end{aligned}[/tex]
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