Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
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Lim X → 1 X 3 + 3 X 2 − 6 X + 2 X 3 + 3 X 2 − 3 X − 1 - Mathematics

\[\lim_{x \to 1} \frac{x^3 + 3 x^2 - 6x + 2}{x^3 + 3 x^2 - 3x - 1}\]

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Solution

Let p(x) = x3 + 3x2 \[-\] 6x + 2
p(1) = 1 + 3\[-\] 6 + 2
       = 0

Now,

\[\left( x + 2 \right)\] is a factor of p(x). 

p(x) = (x\[-\] 1)(x2 + 4x \[-\] 2)

q(x) = x3 + 3x2
\[-\]3x + 2
q(1) = 1 + 3 \[-\]3 \[-\]1
       = 0

\[Now, \left( x + 2 \right)\] is a factor of p(x).

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

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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 29 Limits
Exercise 29.3 | Q 30 | Page 23
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