Find $\,\lim_{x\to 1}\frac{\sqrt[n]{x}-1}{\sqrt[m]{x}-1}$ How do I calculate $\lim_{x\to 1}\frac{\sqrt[n]{x}-1}{\sqrt[m]{x}-1}$.
Please help me.
Thanks!
 A: What about a little l'Hospital?
$$\lim_{x\to 1}\frac{x^{1/n}-1}{x^{1/m}-1}=\lim_{x\to 1}\frac{\frac1nx^{1/n -1}}{\frac1mx^{1/m-1}}=\lim_{x\to 1}\frac mn x^{\frac1n-\frac1m}=\frac mn$$
A: Note that, as $x\to 1$, 
$$
\frac{\sqrt[n]{x}-1}{\sqrt[m]{x}-1}=\frac{\frac{\sqrt[n]{x}-1}{x-1}}{\frac{\sqrt[m]{x}-1}{x-1}}=\frac{\frac{f(x)-f(1)}{x-1}}{\frac{g(x)-g(1)}{x-1}}
\longrightarrow\frac{f'(1)}{g'(1)},
$$
where
$$
f(x)=\sqrt[n]{x}, \quad g(x)=\sqrt[m]{x}.
$$
And as
$$
f'(x)=\big(\sqrt[n]{x}\big)'=\frac{1}{nx^{1-1/n}}, \quad\text{and}\quad f'(1)=\frac{1}{n},
$$
then
$$
\frac{\sqrt[n]{x}-1}{\sqrt[m]{x}-1}=\frac{\frac{\sqrt[n]{x}-1}{x-1}}{\frac{\sqrt[m]{x}-1}{x-1}}\to\frac{m}{n}.
$$
A: You may write 
$$\frac{\sqrt[n]{x}-1}{\sqrt[m]{x}-1}=\frac{e^{\frac{\ln x}{n}}-1}{e^{\frac{\ln x}{m}}-1}$$ then, as $x$ tends to $1$, you have $$e^{\frac{\ln x}{n}}= 1+\frac{\ln x}{n}+o\left(\frac{\ln x}{n}\right)$$
$$e^{\frac{\ln x}{m}}= 1+\frac{\ln x}{m}+o\left(\frac{\ln x}{m}\right)$$ finally you get 
$$
\frac{\frac{\ln x}{n}+o\left(\frac{\ln x}{n}\right)}{\frac{\ln x}{m}+o\left(\frac{\ln x}{n}\right)}=\ldots
$$ and you can easily conclude.
A: We assume that $m$ and $n$ are positive integers. Let $x=y^{mn}$. We are looking for
$$\lim_{y\to 1} \frac{y^m-1}{y^n-1}.\tag{1}$$
Dividing top and bottom by $y-1$, we find that the limit (1) is equal to
$$\lim_{y\to 1} \frac{y^{m-1}+y^{m-2}+\cdots +1}{y^{n-1}+y^{n-2}+\cdots +1},$$
which is $\frac{m}{n}$.  
