Pertanyaan

x → 1 lim x 2 + 4 x − 5 ( 3 x − 1 ) 2 − 4 = ...

$x \to 1 lim \frac{( 3 x - 1 ) ^{2} - 4}{x ^{2} + 4 x - 5} =$ ...

E. Dwi

Master Teacher

Mahasiswa/Alumni Universitas Sriwijaya

Jawaban terverifikasi

Jawaban

nilai dari adalah .

nilai dari $limit as x rightwards arrow 1 of space fraction numerator open parentheses 3 x minus 1 close parentheses squared minus 4 over denominator x squared plus 4 x minus 5 end fraction$ adalah

Pembahasan

Nilai dari : Jadi, nilai dari adalah .

Nilai dari $limit as x rightwards arrow 1 of space fraction numerator open parentheses 3 x minus 1 close parentheses squared minus 4 over denominator x squared plus 4 x minus 5 end fraction$ :

$table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 1 of space fraction numerator open parentheses 3 x minus 1 close parentheses squared minus 4 over denominator x squared plus 4 x minus 5 end fraction end cell equals cell limit as x rightwards arrow 1 of space fraction numerator 9 x squared minus 6 x plus 1 minus 4 over denominator x squared plus 4 x minus 5 end fraction end cell row blank equals cell limit as x rightwards arrow 1 of space fraction numerator 9 x squared minus 6 x minus 3 over denominator x squared plus 4 x minus 5 end fraction end cell row blank equals cell limit as x rightwards arrow 1 of space fraction numerator open parentheses 9 x plus 3 close parentheses open parentheses x minus 1 close parentheses over denominator open parentheses x plus 5 close parentheses open parentheses x minus 1 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 1 of space fraction numerator open parentheses 9 x plus 3 close parentheses up diagonal strike open parentheses x minus 1 close parentheses end strike over denominator open parentheses x plus 5 close parentheses up diagonal strike open parentheses x minus 1 close parentheses end strike end fraction end cell row blank equals cell limit as x rightwards arrow 1 of space fraction numerator open parentheses 9 x plus 3 close parentheses over denominator open parentheses x plus 5 close parentheses end fraction end cell row blank equals cell fraction numerator 9 open parentheses 1 close parentheses plus 3 over denominator 1 plus 5 end fraction end cell row blank equals cell fraction numerator 9 plus 3 over denominator 6 end fraction end cell row blank equals cell 12 over 6 end cell row blank equals 2 end table$

Jadi, nilai dari $limit as x rightwards arrow 1 of space fraction numerator open parentheses 3 x minus 1 close parentheses squared minus 4 over denominator x squared plus 4 x minus 5 end fraction$ adalah .