SMA

Nilai x→1limx3−1×2−5x+4=…

AdministratorMarch 2, 2022

0 277 Less than a minute

Nilai

Pembahasan

Nilai $limit as x rightwards arrow 1 of fraction numerator x squared minus 5 x plus 4 over denominator x cubed minus 1 end fraction$ dapat ditentukan dengan mensubstitusi nilai x equals 1 ke dalam fungsi $fraction numerator x squared minus 5 x plus 4 over denominator x cubed minus 1 end fraction$ sehingga

$table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 1 of fraction numerator x squared minus 5 x plus 4 over denominator x cubed minus 1 end fraction end cell equals cell fraction numerator open parentheses 1 close parentheses squared minus 5 open parentheses 1 close parentheses plus 4 over denominator open parentheses 1 close parentheses cubed minus 1 end fraction end cell row cell limit as x rightwards arrow 1 of fraction numerator x squared minus 5 x plus 4 over denominator x cubed minus 1 end fraction end cell equals cell fraction numerator 1 minus 5 plus 4 over denominator 1 minus 1 end fraction end cell row cell limit as x rightwards arrow 1 of fraction numerator x squared minus 5 x plus 4 over denominator x cubed minus 1 end fraction end cell equals cell 0 over 0 end cell end table$

Karena hasil $limit as x rightwards arrow 1 of fraction numerator x squared minus 5 x plus 4 over denominator x cubed minus 1 end fraction$ menggunakan konsep limit yaitu cara substitusi adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 0 over 0 comma end cell end table maka untuk menentukan limitnya menggunakan cara selanjutnya yaitu pemfaktoran sebagai berikut

$table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 1 of fraction numerator x squared minus 5 x plus 4 over denominator x cubed minus 1 end fraction end cell equals cell limit as x rightwards arrow 1 of fraction numerator up diagonal strike open parentheses x minus 1 close parentheses end strike open parentheses x minus 4 close parentheses over denominator up diagonal strike open parentheses x minus 1 close parentheses end strike open parentheses x squared plus x plus 1 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 1 of fraction numerator x minus 4 over denominator x squared plus x plus 1 end fraction end cell row blank equals cell fraction numerator 1 minus 4 over denominator open parentheses 1 close parentheses squared plus 1 plus 1 end fraction end cell row blank equals cell fraction numerator negative 3 over denominator 3 end fraction end cell row blank equals cell negative 1 end cell end table$

Jadi nilai $limit as x rightwards arrow 1 of fraction numerator x squared minus 5 x plus 4 over denominator x cubed minus 1 end fraction equals negative 1.$

AdministratorMarch 2, 2022

0 277 Less than a minute