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WAEC Further Mathematics 2022 Paper

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Question : 43

Total: 45

Evaluate: lim

_{x\to -2}

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

.

12
8
4
2

Solution:

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

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)

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

x = -2

(-2)

^{2}

- 2(2) + 4 = 4 - 4 + 4

= 4

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