I have this limit to find:
$\lim\limits_{x\to0}\frac{\sqrt[n]{1+x}-1}{x}$
My development was:
Let $\large{u^n = 1+x}$, from here if $x\to 0$ implies that $\large{u^n \to 1}$
And i got: $\Large{\lim_{u^n \to 1}\frac{u-1}{u^n - 1}}$ and using that $\Large{u^n - 1 = (u-1)\sum_{j=0}^{n-1}{u^j}}$
Finally i got $\Large{\lim_{u^n\to1}\frac{1}{\sum_{j=0}^{n-1}{u^j}} = \dfrac{1}{n}}$
I know that the result is correct, but i want to know if all my steps are correct.
Thanks in advance.
Your steps are correct except you write $u\to 1$ instead of $u^n\to 1$. Here is another simple approach $$\lim_{x\to 0}\frac{\sqrt[n]{1+x}-1}{x}$$ $$=\lim_{x\to 0}\frac{(1+x)^{1/n}-1}{x}$$ $$=\lim_{x\to 0}\frac{\left(1+\frac{1}{n}\cdot x+\left(\frac1n\right)\left(\frac1n-1\right)\frac{x^2}{2!}+\ldots\right)-1}{x}$$ $$=\lim_{x\to 0}\left(\frac{1}{n}+\frac{(1-n)x}{2n^2}+\ldots\right)$$ $$=\frac1n$$