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

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

$\lim_{x \to 1} \frac{\left( 2x - 3 \right) \left( \sqrt{x} - 1 \right)}{3 x^2 + 3x - 6}$

#### Solution

$\lim_{x \to 1} \left[ \frac{\left( 2x - 3 \right)\left( \sqrt{x} - 1 \right)}{3 x^2 + 3x - 6} \right]$ It is of the form $\frac{0}{0}$

⇒  $\lim_{x \to 1} \left[ \frac{\left( 2x - 3 \right)\left( \sqrt{x} - 1 \right)}{3\left( x^2 + x - 2 \right)} \right]$

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

=  $\lim_{x \to 1} \left[ \frac{\left( 2x - 3 \right)\left( \sqrt{x} - 1 \right)}{3\left( x\left( x + 2 \right) - 1\left( x + 2 \right) \right)} \right]$

=  $\lim_{x \to 1} \left[ \frac{\left( 2x - 3 \right)\left( \sqrt{x} - 1 \right)}{3\left( x - 1 \right)\left( x + 2 \right)} \right]$

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

=  $\lim_{x \to 1} \left[ \frac{\left( 2x - 3 \right)\left( \sqrt{x} - 1 \right)}{3\left( \sqrt{x} + 1 \right)\left( \sqrt{x} - 1 \right)\left( x + 2 \right)} \right]$

=  $\frac{- 1}{3\left( 2 \right) \times 3}$

=  $\frac{- 1}{18}$

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 29 Limits
Exercise 29.4 | Q 28 | Page 29