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WAEC Further Mathematics 2016 Paper

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Question : 26

Total: 40

Find

\lim\limits_{x \to 3} \frac{2x^{2}+x-21}{x-3}

.

0
1
7
13

Solution:

Concept:

Substituting

x=3

directly gives

\frac{0}{0}

, so factorise the numerator first and cancel the common factor.

Explanation:

Factorise

2x^{2}+x-21

.

We need two numbers whose product is

2 \times (-21) = -42

and whose sum is

1

.

These numbers are

7

and

-6

.

So,

2x^{2}+x-21 = 2x^{2}-6x+7x-21

.

Group terms:

2x(x-3) + 7(x-3) = (2x+7)(x-3)

.

Thus,

\frac{2x^{2}+x-21}{x-3} = \frac{(2x+7)(x-3)}{x-3}

.

Since

x \to 3

, we have

x \ne 3

, so cancel

x-3

.

The required limit becomes

\lim\limits_{x \to 3} (2x+7)

.

Substitute

x=3

:

2(3)+7 = 6+7 = 13

.

Answer:

D. 13

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