Test Index

WAEC Further Mathematics 2022 Paper

© africaexams.com

Question : 43

Total: 45

Evaluate: lim

_{x\to -2}

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

.

12
8
4
2

Solution:

Concept:

Factorise the numerator using the sum of cubes identity, cancel the common factor, then substitute

x=-2

.

Explanation:

Write

8

as

2^3

, so

x^3+8=x^3+2^3

.

Use the identity:

x^3+y^3=(x+y)(x^2-xy+y^2)

.

Here

y=2

, so

x^3+8=(x+2)(x^2-2x+4)

.

Therefore,

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

for

x\neq -2

.

Now substitute

x=-2

:

(-2)^2-2(-2)+4=4+4+4=12

.

Answer:

12

(Option A).

© africaexams.com

Go to Question:

Prev Question

Next Question