# What is lim_(x->0) ((1+x)^n-1)/x ?

Nov 4, 2017

${\lim}_{x \to 0} \frac{{\left(1 + x\right)}^{n} - 1}{x} = n$

#### Explanation:

If we assume the question was: ${\lim}_{x \to 0} \frac{{\left(1 + x\right)}^{n} - 1}{x}$

We have an indeterminate limit of the form $\frac{0}{0}$

Hence, L'Hopital's rule applies.

$\therefore {\lim}_{x \to 0} \frac{{\left(1 + x\right)}^{n} - 1}{x} = {\lim}_{x \to 0} \frac{\frac{d}{\mathrm{dx}} \left({\left(1 + x\right)}^{n} - 1\right)}{\frac{d}{\mathrm{dx}} \left(x\right)}$

$= {\lim}_{x \to 0} \frac{n {\left(1 + x\right)}^{n - 1} \times 1 - 0}{1}$ [Power rule and chain rule]

$= {\lim}_{x \to 0} n {\left(1 + x\right)}^{n - 1}$

$= n \times {1}^{n - 1} = n$