Question

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

Answer 1

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

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

\[\Rightarrow\]y ​→ 1 

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