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What methods can be used to evaluate the limit: $$\lim_{x\to 0} \frac{\sqrt[n]{1+x}-1}{x}, n \in \Bbb Z$$

By the way, as a rule, I use method with conjugate expression for removing problem like this $$ \sqrt[]a - \sqrt[]b = \frac{(\sqrt[]a - \sqrt[]b)(\sqrt[]a + \sqrt[]b)}{ \sqrt[]a + \sqrt[]b} = \frac{a-b}{\sqrt[]a+\sqrt[]b}$$

but I don't know how to evaluate it for nth root. Maybe, this issue can be solve by using mathematical induction method, but I have not right outcome. And, yes, I have heard about L'Hopital rule.

Perfectionist
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    Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Shaun Nov 17 '21 at 17:21
  • Power series for nth root. – herb steinberg Nov 17 '21 at 17:36
  • You need to say what do you know about limits, there are several ways to do it depending on your knowledge on taylor series, derivatives or other tools. – Marcos Nov 17 '21 at 17:42
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    Have you heard of L'Hopital? – finmathstudent Nov 17 '21 at 17:59
  • @Shaun I have edited my question. Is it enough? – Perfectionist Nov 17 '21 at 18:29
  • That's much better, thank you. – Shaun Nov 17 '21 at 18:30
  • Hint: $a^n-1=(a-1)(a^{n-1}+a^{n-2}+\cdots+1)$, so $$a-1={a^n-1\over a^{n-1}+a^{n-2}+\cdots+1}$$ Now let $a=\sqrt[n]{1+x}$. – Barry Cipra Nov 17 '21 at 18:36

6 Answers6

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You may use

$$\sqrt[n]a-\sqrt[n]b = \frac{a-b}{\sqrt[n]{a^{n-1}}+\sqrt[n]{a^{n-2}b}+\cdots+\sqrt[n]{b^{n-1}}}$$

to mimic the case of $n=2$.

jjagmath
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Here's another, less elementary approach, which may be useful to some.

Recall the definition of the derivative: $$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}h.$$ Using this, your limit is $$\frac d{dx}\sqrt[n]{1+x}\bigg|_{x=0}.$$ We have $$\frac {d(1+x)^{1/n}}{dx}=\frac1n(1+x)^{1/n-1},$$ so the derivative at $0$ is $1/n$.

Carl Schildkraut
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    Calculating the derivative of $\sqrt[n]{1+x}$ involves a similar limit (actually, one that is a little more difficult), so this reasoning is circular. – jjagmath Nov 17 '21 at 18:57
  • @jjagmath This being circular depends on what you’re allowed to assume. – Dionel Jaime Nov 17 '21 at 19:33
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    @jjagmath That's a good point, but I think this is frequently how people are taught to evaluate limits such as these. In any case, I think the intuition that "this limit looks like a derivative" is good to showcase, even if this solution doesn't make as much sense from a "mathematics from first principles" point of view. – Carl Schildkraut Nov 17 '21 at 19:45
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$$\lim_{x\to 0} \frac{\sqrt[n]{1+x}-1}{x}, n \in \Bbb Z$$

Let's work with numerator. Thanks to the answers of my colleagues I will use this one:

$$\sqrt[n]{1+x}-1=\sqrt[n]{1+x}-\sqrt[n]{1} = \frac{1+x -1}{\sqrt[n]{{(1+x)^n}{1^0}}+\sqrt[n]{(1+x)^{n-1}}{1^1}+...+\sqrt[n]{{(1+x)^0}{1^{n-1}}}}=\frac{x}{1+1+...+1} $$ We have power from 0 to n-1, so it will be n number of 1: $$\frac{x}{n}$$ So, the last step to enter this into the numerator and evaluate: $$\lim_{x\to 0} \frac{\frac{x}{n}}{x}=\lim_{x\to 0} \frac{x}{nx}=\frac{1}{n}$$

Perfectionist
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  • The best elementary approach is in my community wiki answer because it generalizes to $$\lim_{x\to 0}\frac{(1+x)^a-1}x=a\in\Bbb R$$ where your method fails. – PinkyWay Nov 17 '21 at 21:08
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$$y=\frac{\sqrt[n]{1+x}-1}{x}=\frac{A-1}{x}$$ $$A=\sqrt[n]{1+x}\implies \log(A)=\frac 1n \log(1+x)=\frac 1n \Big[x-\frac{x^2}{2}+O\left(x^3\right) \Big]$$ $$A=e^{\log(A)}=1+\frac{x}{n}-\frac{(n-1) x^2}{2 n^2}+O\left(x^3\right)$$ $$y=\frac{A-1}{x}=\frac{1}{n}-\frac{(n-1) x}{2 n^2}+O\left(x^2\right)$$ which shows the limit but also how it is approached.

Claude Leibovici
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$$\lim_{x\to 0}\frac{\sqrt[n]{1+x}-1}x=\lim_{x\to 0}\frac{e^{\frac1n\ln(1+x)}-1}x=\lim_{x\to0}\frac{e^{\frac1n\ln(1+x)}-1}{\frac1n\ln(1+x)}\cdot\frac1n\frac{\ln(1+x)}x=\frac1n.$$ Just use the manual limits $$\boxed{\lim_{x\to 0}\frac{e^x-1}x=1}\\\boxed{\lim_{x\to 0}\frac{\ln(1+x)}x=1}$$

PinkyWay
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Well, I have an interesting approach to it using non standard Analysis. The limit is nothing but, $$st\left(\frac{(1+dx)^{1/n} - 1}{dx}\right)$$ According to chapter 17 of Euler's book on "The analysis on infinity" $1+dx=e^{dx}$. $$st\left(\frac{e^{\frac{dx}{n}}-1}{dx}\right)$$ In the same chapter using binomial theorem, Euler managed to prove the series expansion of $e^{x}$ Replacing $x$ with $\frac{dx}{n}$ we have that, $$st\left(\frac{1}{dx}\sum_{k\ge 1}\frac{(dx/n)^{k}}{k!}\right)$$ $1$ cancels off each other, and power of $dx$ is decreased by $1$ as we are dividing by it. $$st\left(\frac{1}{n}+(...)dx+(....)(dx)^{2}+...\right)$$ By definition of standard part function we have that it's $\frac{1}{n}$. So, $$\boxed{\lim_{x\to 0}\frac{\sqrt[n]{1+x}-1}{x}=\frac{1}{n}}$$

2 is even prime
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