Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
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Lim X → − 2 X 3 + X 2 + 4 X + 12 X 3 − 3 X + 2 - Mathematics

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

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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
       = (+ 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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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 29 Limits
Exercise 29.3 | Q 29 | Page 23
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