Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# Lim X → 1 X 3 + 3 X 2 − 6 X + 2 X 3 + 3 X 2 − 3 X − 1 - Mathematics

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

#### 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).

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

Is there an error in this question or solution?

#### APPEARS IN

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