Find $\,\lim_{x\to 1}\frac{\sqrt[n]{x}-1}{\sqrt[m]{x}-1}$

Question

How do I calculate $\lim_{x\to 1}\frac{\sqrt[n]{x}-1}{\sqrt[m]{x}-1}$.

Please help me.

Thanks!

Answer 1

What about a little l'Hospital?

$$\lim_{x\to 1}\frac{x^{1/n}-1}{x^{1/m}-1}=\lim_{x\to 1}\frac{\frac1nx^{1/n -1}}{\frac1mx^{1/m-1}}=\lim_{x\to 1}\frac mn x^{\frac1n-\frac1m}=\frac mn$$

Answer 2

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 $$ \frac{\sqrt[n]{x}-1}{\sqrt[m]{x}-1}=\frac{\frac{\sqrt[n]{x}-1}{x-1}}{\frac{\sqrt[m]{x}-1}{x-1}}\to\frac{m}{n}. $$

Answer 3

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.

Answer 4

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 $\frac{m}{n}$.