Limits: How to evaluate $\lim_{x\to 0}\frac{\sqrt[n]{1+x}-1}{x},n\in\Bbb Z$ 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.
 A: 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 $n=2$.
A: 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 $1/n$.
A: $$\lim_{x\to 0} \frac{\sqrt[n]{1+x}-1}{x}, n \in \Bbb Z$$
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:
$$\lim_{x\to 0} \frac{\frac{x}{n}}{x}=\lim_{x\to 0} \frac{x}{nx}=\frac{1}{n}$$
A: $$y=\frac{\sqrt[n]{1+x}-1}{x}=\frac{A-1}{x}$$
$$A=\sqrt[n]{1+x}\implies \log(A)=\frac 1n \log(1+x)=\frac 1n \Big[x-\frac{x^2}{2}+O\left(x^3\right) \Big]$$
$$A=e^{\log(A)}=1+\frac{x}{n}-\frac{(n-1) x^2}{2 n^2}+O\left(x^3\right)$$
$$y=\frac{A-1}{x}=\frac{1}{n}-\frac{(n-1) x}{2 n^2}+O\left(x^2\right)$$ which shows the limit but also how it is approached.
A: $$\lim_{x\to 0}\frac{\sqrt[n]{1+x}-1}x=\lim_{x\to 0}\frac{e^{\frac1n\ln(1+x)}-1}x=\lim_{x\to0}\frac{e^{\frac1n\ln(1+x)}-1}{\frac1n\ln(1+x)}\cdot\frac1n\frac{\ln(1+x)}x=\frac1n.$$
Just use the manual limits $$\boxed{\lim_{x\to 0}\frac{e^x-1}x=1}\\\boxed{\lim_{x\to 0}\frac{\ln(1+x)}x=1}$$
A: 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,
$$\boxed{\lim_{x\to 0}\frac{\sqrt[n]{1+x}-1}{x}=\frac{1}{n}}$$
