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Question: If \(f(x) = \frac{2}{x - 3},g(x) = \frac{x - 3}{x + 4}\) and \(h(x) = - \frac{2(2x + 1)}{x^{2} + x -...

If f(x)=2x−3,g(x)=x−3x+4f(x) = \frac{2}{x - 3},g(x) = \frac{x - 3}{x + 4} and h(x)=−2(2x+1)x2+x−12h(x) = - \frac{2(2x + 1)}{x^{2} + x - 12}then

lim⁡x→3[f(x)+g(x)+h(x)]\lim_{x \rightarrow 3}\lbrack f(x) + g(x) + h(x)\rbrack is

A

– 2

B

– 1

C

−27- \frac{2}{7}

D

0

Answer

−27- \frac{2}{7}

Explanation

Solution

We have f(x)+g(x)+h(x)=x2−4x+17−4x−2x2+x−12f(x) + g(x) + h(x) = \frac{x^{2} - 4x + 17 - 4x - 2}{x^{2} + x - 12}

=x2−8x+15x2+x−12=(x−3)(x−5)(x−3)(x+4)= \frac{x^{2} - 8x + 15}{x^{2} + x - 12} = \frac{(x - 3)(x - 5)}{(x - 3)(x + 4)}

∴lim⁡x→3[f(x)+g(x)+h(x)]=lim⁡x→3(x−3)(x−5)(x−3)(x+4)=−27\therefore\lim_{x \rightarrow 3}\lbrack f(x) + g(x) + h(x)\rbrack = \lim_{x \rightarrow 3}\frac{(x - 3)(x - 5)}{(x - 3)(x + 4)} = - \frac{2}{7}.