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

Assessment: WAEC SSCE - Further Mathematics - 2022 Subject: Further Mathematics

Question 1 Report

Answer

$\frac{x^{3} + 8}{x + 2}$

x $^{3}$ + 8 = x $^{3}$ + 23
recall that
x $^{3}$ + y $^{3}$ = (x+y)(x $^{2}$ - xy + y $^{2}$ )
x = x, y = 2
x $^{3}$ + 8 = (x+2)(x $^{3}$ - 2(x) + 22)

$\frac{x^{3} + 8}{x + 2} = \frac{(x + 2) (x^{3} - 2 (x) + 22)}{x + 2}$ - 2x + 4

x = -2
(-2) $^{2}$ - 2(2) + 4 = 4 - 4 + 4

= 4

Read lesson note on Differentiation (WAEC)

Answer Details

$\frac{x^{3} + 8}{x + 2}$

x $^{3}$ + 8 = x $^{3}$ + 23
recall that
x $^{3}$ + y $^{3}$ = (x+y)(x $^{2}$ - xy + y $^{2}$ )
x = x, y = 2
x $^{3}$ + 8 = (x+2)(x $^{3}$ - 2(x) + 22)

$\frac{x^{3} + 8}{x + 2} = \frac{(x + 2) (x^{3} - 2 (x) + 22)}{x + 2}$ - 2x + 4

x = -2
(-2) $^{2}$ - 2(2) + 4 = 4 - 4 + 4

= 4

Read lesson note on Differentiation (WAEC)

Differentiation

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

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

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