Question

\[\lim_{x \to - 2} \frac{x^3 + x^2 + 4x + 12}{x^3 - 3x + 2}\]

Answer 1

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)

\[\Rightarrow \lim_{x \to - 2} \left[ \frac{x^3 + x^2 + 4x + 12}{x^3 - 3x + 2} \right]\]
\[ = \lim_{x \to - 2} \left[ \frac{\left( x + 2 \right)\left( x^2 - x + 6 \right)}{\left( x + 2 \right)\left( x^2 - 2x + 1 \right)} \right]\]
\[ = \frac{(  2 )^2 - \left( - 2 \right) + 6}{\left( - 2 \right)^2 - 2\left( - 2 \right) + 1}\]
\[ = \frac{4 + 2 + 6}{4 + 4 + 1}\]
\[ = \frac{12}{9}\]
\[ = \frac{4}{3}\]