Question

Hasil dari
[tex] \displaystyle \lim_{x \to2} \frac{ {x}^{2} + 2x - 8}{ {x}^{3} - 12x} [/tex]
adalah ...

#Opsinya blm dibeli ​

Answer 1

Verified answer

Penjelasan dengan langkah-langkah:

limit

[tex] lim_ { \: x \: \to \: 2} \: \frac{ {x}^{2} + 2x - 8 }{ {x}^{3} - 12x} = \frac{(x + 4)(x - 2)}{x( {x}^{2} - 12) } \\ [/tex]

substitusi nilai x = 2

[tex] \frac{(x + 4)(x - 2)}{x( {x}^{2} - 12)} = \frac{(2 + 4)(2 - 2)}{2( {2}^{2} - 12)} \\ \frac{(8)(0)}{2( - 8)} = - \frac{0}{16} = 0[/tex]

Jadi, hasil dari [tex] \rm lim_ { \: x \: \to \: 2} \: \frac{ {x}^{2} + 2x - 8 }{ {x}^{3} - 12x} \\ [/tex] adalah 0 .

Answer 2

Limit

[Metode L'Hopital]

..

[tex]\begin{aligned} \displaystyle \lim_{x \to2}& \left( \frac{ {x}^{2} + 2x - 8}{ {x}^{3} - 12x} \right) \\ \displaystyle \lim_{x \to2}& \left( \frac{\frac{d}{dx}(x^2 + 2x - 8)}{\frac{d}{dx}(x^3 - 12x)} \right) \\ \displaystyle \lim_{x \to2}& \left( \frac{2x + 2}{3x^2 - 12} \right) \end{aligned}[/tex]

[tex]\begin{aligned} &= \frac{2(2) + 3}{3(2)^2 - 12} \\&= \frac{4 + 3}{12 - 12} \\&= \frac{7}{0} \\&= \boxed{\bold{\underline{Tidak~Terdefinisi}}} \end{aligned}[/tex]

[tex]\begin{array}{lr}\texttt{Memperingati Hari Perawat Internasional}\end{array}[/tex]

[tex]\boxed{\colorbox{ccddff}{Answered by Danial Alf'at | 12 - 05 - 2023}}[/tex]