Lim X → 2 X 3 + 3 X 2 − 9 X − 2 X 3 − X − 6 - Mathematics

$\lim_{x \to 2} \frac{x^3 + 3 x^2 - 9x - 2}{x^3 - x - 6}$

Solution

$\lim_{x \to 2} \left[ \frac{x^3 + 3 x^2 - 9x - 2}{x^3 - x - 6} \right]$ It is of the form $\frac{0}{0} .$

Let p(x) = x3 + 3x$-$9x$-$ 2
p(2) = 8 + 12 $-$18 $-$2
= 0

Now,

$\left( x - 2 \right)$ is a factor of p(x).

Let q(x) = x3 – x – 6
q(2) = 8 – 2 – 6
= 0

$Now, \left( x - 2 \right)$is a factor of q(x).

$\Rightarrow \lim_{x \to 2} \left[ \frac{x^3 + 3 x^2 - 9x - 2}{x^3 - x - 6} \right]$
$= \lim_{x \to 2} \left[ \frac{\left( x - 2 \right)\left( x^2 + 5x + 1 \right)}{\left( x - 2 \right)\left( x^2 + 2x + 3 \right)} \right]$
$= \lim_{x \to 2} \left[ \frac{x^2 + 5x + 1}{x^2 + 2x + 3} \right]$
$= \frac{(2 )^2 + 5 \times 2 + 1}{\left( 2 \right)^2 + 2 \times 2 + 3}$
$= \frac{4 + 10 + 1}{4 + 4 + 3}$
$= \frac{15}{11}$

Is there an error in this question or solution?

APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 29 Limits
Exercise 29.3 | Q 26 | Page 23