Is my limit development correct?

Question

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 work is correct. You could make it shorter using the binomial expansion $\sqrt[n] {1+x}=1+\frac{x}{n}+\frac{(1-n) x^2}{2 n^2}+O\left(x^3\right)$ which will show that the expression is $\frac{1}{n}+\frac{(1-n) x}{2 n^2}+O\left(x^2\right)$ which shows the limit and also how it is approached. — Claude Leibovici, May 27 '20 at 04:51
Well you just write $u\to 1$ instead of $u^n\to 1$. Rest is fine. — Paramanand Singh, May 27 '20 at 05:17
@ParamanandSingh Why write $\large{u\to 1}$ instead of $\large{u^n \to 1}$? Is a convention or what? — ESCM, May 27 '20 at 16:25
The limit notation is $\lim\limits_{\text{variable} \to\text{value} } $. Here $u$ is a variable and $u^n$ is a function of the variable $u$. The limit notation does not allow for $\lim\limits_{\text{function} \to\text{value}} $. — Paramanand Singh, May 27 '20 at 16:43

Harish Chandra Rajpoot · Accepted Answer · 2020-05-27T05:22:06.463

1

Your steps are correct except you write $u\to 1$ instead of $u^n\to 1$. Here is another simple approach $$\lim_{x\to 0}\frac{\sqrt[n]{1+x}-1}{x}$$ $$=\lim_{x\to 0}\frac{(1+x)^{1/n}-1}{x}$$ $$=\lim_{x\to 0}\frac{\left(1+\frac{1}{n}\cdot x+\left(\frac1n\right)\left(\frac1n-1\right)\frac{x^2}{2!}+\ldots\right)-1}{x}$$ $$=\lim_{x\to 0}\left(\frac{1}{n}+\frac{(1-n)x}{2n^2}+\ldots\right)$$ $$=\frac1n$$

edited May 27 '20 at 05:22

answered May 27 '20 at 05:16

Harish Chandra Rajpoot

37,450

Why write $\large{u\to 1}$ instead of $\large{u^n \to 1}$? Is a convention or what? – ESCM May 27 '20 at 16:25
well, you have have assumed a new variable $u$ not $u^{n}$ so you should express limit for $u$. – Harish Chandra Rajpoot May 27 '20 at 17:13

Is my limit development correct?

1 Answers1