Evaluate: \( \lim_{x \to -2} \frac{x^3 + 8}{x + 2} \).

Assessment: WAEC SSCE - Further Mathematics - 2022 (Essay) Subject: Further Mathematics

Question 1 Report

Evaluate: \( \lim_{x \to -2} \frac{x^3 + 8}{x + 2} \).

Answer Details

x3+8x+2

x3
 + 8 = x3
 + 23
recall that
x3 3  + y3 3  = (x+y)(x2 2  - xy + y2 2 )
x = x, y = 2
x3 3  + 8 = (x+2)(x3 3  - 2(x) + 22)

x3+8x+2=(x+2)(x3−2(x)+22)x+2
 - 2x + 4

x = -2
(-2)2
 - 2(2) + 4 = 4 - 4 + 4

= 4

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