Question

Limit fungsi aljabar (faktorisasi)

x = 2x^2 -7x + 3 / x^2 + x - 12
x -> 3

note:
sebelumnya sy cari faktorisasi 2x^2 - 7x + 3 itu sy hitung dapat sampai 2x (x - 3) - 1 (x + 3) tapi - sama + . tolong we gw kurang faham.​ ini soalnya salah atau saya

Answer 1

Verified answer

Jawaban:

[tex] \displaystyle \lim_{x \to3} \frac{2 {x}^{2} - 7x + 3 }{ {x}^{2} + x - 12} = \bold{\frac{5}{7} }[/tex]

Penjelasan dengan langkah-langkah:

nih sy kasih tau

[tex] \displaystyle \lim_{x \to3} \frac{2 {x}^{2} - 7x + 3 }{ {x}^{2} + x - 12} [/tex]

2x² - 7x + 3 difaktorkan:

= 2x² - 6x - x + 3

= (2x - 1)(x - 3)

x² + x - 12 difaktorkan:

= x² + 4x - 3x - 12

= (x + 4)(x - 3)

Jadi...

[tex] \displaystyle \lim_{x \to3} \frac{2 {x}^{2} - 7x + 3 }{ {x}^{2} + x - 12} [/tex]

[tex] = \displaystyle \lim_{x \to3} \frac{(2x - 1)(x - 3) }{ (x + 4)(x - 3)} [/tex]

[tex] = \displaystyle \lim_{x \to3} \frac{(2x - 1) \cancel{(x - 3) }}{ (x + 4) \cancel{(x - 3)}} [/tex]

[tex] = \displaystyle \lim_{x \to3} \frac{2x - 1 }{ x + 4} [/tex]

[tex] = \displaystyle \frac{2(3) - 1}{3 + 4} [/tex]

[tex] = \displaystyle \frac{6 - 1}{7} [/tex]

[tex] = \displaystyle \frac{5}{7} [/tex]