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Question
\[\lim_{x \to - 2} \frac{x^3 + x^2 + 4x + 12}{x^3 - 3x + 2}\]
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Solution
Let p(x) = x3 + x2 + 4x + 12
p(–2) = 0
Thus, x = –2 is the root of p(x).
Now,
\[\left( x + 2 \right)\] is a factor of p(x).
p(x) = x3 + x2 + 4x + 12
= (x + 2)(x2 – x + 6)
Let q(x) = x3 – 3x + 2
q\[\left( - 2 \right)\] =\[-\]8 + 6 + 2
= 0
Thus, x =\[-\]2 is the root of q(x).
Now, \[\left( x + 2 \right)\]is a factor of q(x).
q(x) = (x + 2)(x2 – 2x + 1)
\[\Rightarrow \lim_{x \to - 2} \left[ \frac{x^3 + x^2 + 4x + 12}{x^3 - 3x + 2} \right]\]
\[ = \lim_{x \to - 2} \left[ \frac{\left( x + 2 \right)\left( x^2 - x + 6 \right)}{\left( x + 2 \right)\left( x^2 - 2x + 1 \right)} \right]\]
\[ = \frac{( 2 )^2 - \left( - 2 \right) + 6}{\left( - 2 \right)^2 - 2\left( - 2 \right) + 1}\]
\[ = \frac{4 + 2 + 6}{4 + 4 + 1}\]
\[ = \frac{12}{9}\]
\[ = \frac{4}{3}\]
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