Math Problem Statement

Tentukan nilai limit berikut (a) lim x→3 x 2−5x+6 x2−4x+3 (b) lim t→−1 t 3−4t 2+t+6 t+1

Solution

Mari kita selesaikan kedua limit berikut:

(a) limx3x25x+6x24x+3\lim_{x \to 3} \frac{x^2 - 5x + 6}{x^2 - 4x + 3}

Langkah-langkah:

  1. Substitusi langsung: Substitusi x=3x = 3 ke dalam fungsi: 325(3)+6324(3)+3=915+6912+3=00\frac{3^2 - 5(3) + 6}{3^2 - 4(3) + 3} = \frac{9 - 15 + 6}{9 - 12 + 3} = \frac{0}{0} Hasilnya adalah bentuk tak tentu 00\frac{0}{0}. Jadi, kita perlu melakukan faktorisasi.

  2. Faktorisasi:

    • Pembilang: x25x+6=(x2)(x3)x^2 - 5x + 6 = (x - 2)(x - 3)
    • Penyebut: x24x+3=(x1)(x3)x^2 - 4x + 3 = (x - 1)(x - 3)

    Maka limit menjadi: [ \lim_{x \to 3} \frac{(x - 2)(x - 3

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Math Problem Analysis

Mathematical Concepts

Limits
Algebra
Factoring

Formulas

Limit of a function as x approaches a value
Factoring quadratic expressions

Theorems

Limit Substitution Theorem
Factoring Theorem for Quadratic Equations

Suitable Grade Level

Grades 10-12