Question

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

Answer 1

$$\begin{gathered} \underset{x\to 0}{\lim }\left[\frac{{\left(1+x\right)}^6-1}{{\left(1+x\right)}^2-1}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{{\left(1+x\right)}^6-1}{x}\times \frac{x}{{\left(1+x\right)}^2-1}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{{\left(1+x\right)}^6-1^6}{\left(1+x\right)-1}\times \frac{\left(1+x\right)-1}{{\left(1+x\right)}^2-1}\right] \end{gathered}$$

Let y = 1 + x

When x ​→ 0, then 1 + x → 1.
$\Rightarrow$y ​→ 1

$$\begin{gathered} \underset{y\to 1}{\lim }\left[\left(\frac{y^6-1^6}{y-1}\right)\times \frac{\left(y-1\right)}{y^2-1^2}\right] \\ =\frac{6\times {\left(1\right)}^{6-1}}{2\times {\left(1\right)}^{2-1}} \\ =\frac{6}{2} \\ =3 \end{gathered}$$