Mathematical Reflections 2016, Issue 2 - Problem U370

Problem:
Evaluate $$\lim_{x \to 0} \dfrac{\sqrt{1+2x}\cdot\sqrt[3]{1+3x}\cdot\ldots\cdot\sqrt[n]{1+nx}-1}{x}.$$

Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam

Solution:
Observe that for any $k=2,\ldots,n$ and for any $x$ such that $|x|<1$, we have $$\sqrt[k]{1+kx}=1+\dfrac{1}{k}kx+o(x^2)=1+x+o(x^2).$$ Hence,
$$\begin{array}{lll}\sqrt{1+2x}\cdot\sqrt[3]{1+3x}\cdot\ldots\cdot\sqrt[n]{1+nx}-1&=&(1+x+o(x^2))^{n-1}-1\\&=&(1+(n-1)x+o(x^2))-1\\&=&(n-1)x+o(x^2). \end{array}$$
Therefore, $$\lim_{x \to 0} \dfrac{\sqrt{1+2x}\cdot\sqrt[3]{1+3x}\cdot\ldots\cdot\sqrt[n]{1+nx}-1}{x}=n-1.$$