Tính L = lim x->0 (căn bậc n (1+ax) - 1)/x , a khác 0

* Hướng dẫn giải

Đáp án A

$\begin{array}{l}L=\underset{x\to 0}{lim}\frac{\sqrt[n]{1+ax}-1}{x}\\ =\underset{x\to 0}{lim}\frac{\left(\sqrt[n]{\left(1+ax\right)}-1\right)\left(1+\sqrt[n]{1+ax}+\sqrt[n]{1+ax}+\mathrm{...}+\sqrt[n]{1+ax}\right)}{x\left(1+\sqrt[n]{\left(1+ax\right)}+\sqrt[n]{{{\left(1+ax\right)}^2}^{}}+\mathrm{...}+\sqrt[n]{{\left(1+ax\right)}^{n-1}}\right)}\\ =\underset{x\to 0}{lim}\frac{1+\text{ax}-1}{x\left(1+\sqrt[n]{1+ax}+\sqrt[n]{1+ax}+\mathrm{...}+\sqrt[n]{1+ax}\right)}=\frac{a}{n}\end{array}$