$\displaystyle\lim_{n \to 0} \dfrac{(1+x)^n-1}{x}\\ =\displaystyle\lim_{n \to 0} \dfrac{(1+x)^n-1^n}{x}\\ =\displaystyle\lim_{n \to 0} \dfrac{(1+x-1)\left((1+x)^{n-1}+(1+x)^{n-2}+\cdots+(1+x)^{0}\right)}{x}\\ =\displaystyle\lim_{n \to 0} (1+x)^{n-1}+(1+x)^{n-2}+\cdots+(1+x)^{0}\\ = \underbrace{1^{n-1}+1^{n-2}+\cdots+1^{0}}_{\text{n số hạng}}\\ =1.n\\ =n$
