# Lim X → 0 ( 1 + X ) 6 − 1 ( 1 + X ) 2 − 1 - Mathematics

$\lim_{x \to 0} \frac{\left( 1 + x \right)^6 - 1}{\left( 1 + x \right)^2 - 1}$

#### Solution

$\lim_{x \to 0} \left[ \frac{\left( 1 + x \right)^6 - 1}{\left( 1 + x \right)^2 - 1} \right]$
$= \lim_{x \to 0} \left[ \frac{\left( 1 + x \right)^6 - 1}{x} \times \frac{x}{\left( 1 + x \right)^2 - 1} \right]$
$= \lim_{x \to 0} \left[ \frac{\left( 1 + x \right)^6 - 1^6}{\left( 1 + x \right) - 1} \times \frac{\left( 1 + x \right) - 1}{\left( 1 + x \right)^2 - 1} \right]$

Let y = 1 + x
When x ​→ 0, then 1 + x → 1.

$\Rightarrow$y ​→ 1

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

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 29 Limits
Exercise 29.5 | Q 3 | Page 33