Finding the limit of $\lim_{x\rightarrow 0}\frac{\sqrt[m]{1+P(x)}-1}{x}$

Question

I need to find the limit of

$\lim_{x\rightarrow 0}\frac{\sqrt[m]{1+P(x)}-1}{x}$

where $P(x) = a_{1}x + a_{2}x^2+...+a_{n}x^n = \sum_{i=1}^{n}a_{i}x^{i}$

My try.

let $n=\frac{1}{m}$

$\lim_{x\rightarrow 0} \frac{\sqrt[m]{1+\sum_{i=1}^{n}a_{i}x^{i}}-1}{x}=\lim_{x\rightarrow 0} \frac{1+n\sum_{i=1}^{n}a_{i}x^{i}+...+(\sum_{i=1}^{n}a_{i}x^{i})^{n}-1}{x}=\lim_{x\rightarrow 0}[n\sum_{i=1}^{n}a_{i}x^{i-1}+...+(\sum_{i=1}^{n}a_{i}x^{i-1})^{n}]=a_{1}nx^{0}+0+0+...+0=a_1n = \frac{a1}{m}$

but I'm not sure whether this solution is right, because I'm not sure if I can expand binomial theorem for the m'th roots?And I would like to see a more elegant solution for this particular problem, if there's any, thank you for your time.

Answer 1

It is the derivative at $0$ of $f(x)=(1+P(x))^{1/m}$

Answer 2

Let $u=\sqrt[m]{1+P(x)}$, we have $$\frac{u-1}{x}=\frac{(u-1)(u^{m-1}+\cdots +u+1)}{x(u^{m-1}+\cdots +u+1)}=\frac{u^m-1}{x(u^{m-1}+\cdots +u+1)}\\=\frac{a_1+\cdots +a_nx^{n-1}}{u^{m-1}+\cdots +u+1}\rightarrow\frac{a_1}{m},\text{ as } x\rightarrow 0,u\rightarrow1.$$