How can you calculate $\lim _{x\rightarrow 1}\; \frac{\root {m} \of {x}-1}{\root {n} \of {x}-1}$without using L'Hospital's law?

Question

I'm trying to calculate this limit with substitution: $\lim _{x\rightarrow 1}\; \frac{\root {m} \of {x}-1}{\root {n} \of {x}-1}$

I started by multiplying with $\frac{\root{n} \of {x}}{\root{n} \of {x}}$ $\to$ $\lim _{x\rightarrow 1}\; \frac{\root {mn} \of {x^{n+m}}-\root{n} \of {x}}{\root {n} \of {x^2}-\root{n} \of {x}}$

I then substituted $\root{mn}\of{x}$ with $t$, so I got $x=t^{mn}$ with $\lim _{x\rightarrow 1}\;\root{mn}\of{x}=1$

$\lim _{t\rightarrow 1}\;\frac{t^{n+m}-\root{n} \of {t^{mn}}}{\root{n} \of {t^{2mn}}-\root{n} \of {t^{mn}}}$ $\to$ $\lim _{t\rightarrow 1}\;\frac{t^{n+m}-t^m}{t^{2n}-t^m}$ $\to$ $\lim _{t\rightarrow 1}\;\frac{t^m(t^n-1)}{t^m(t^m-1)}$ $\to$ $\lim _{t\rightarrow 1}\;\frac{t^n-1}{t^m-1}$

$\lim _{t\rightarrow 1}\;\frac{t^n-1}{t^m-1}$ $\to$ $\lim _{t\rightarrow 1}\;\frac{(t^{n-1}+t^{n-2}+...+t^1+t^0)}{(t^{m-1}+t^{m-2}+...+t^1+t^0)}$

So if we put $1$ in we will get $\frac{n*1}{m*1}$ which is $\frac{n}{m}$

Is this right?

Thanks

Answer 1

I suggest you to use the identity $$ t^N-1=(t-1)\sum_{k=0}^{N-1}t^{k} $$ Hence, by setting $x=t^{mn}$ we obtain that $t\to1$ as $x\to1$. Therefore $$ \lim_{x\to1}\frac{\sqrt[m]{x}-1}{\sqrt[n]{x}-1} =\lim_{t\to1}\frac{\sqrt[m]{t^{mn}}-1}{\sqrt[n]{t^{mn}}-1} =\lim_{t\to1}\frac{t^n-1}{t^m-1}\\ =\lim_{t\to1}\frac{\sum_{k=0}^{n-1}t^{k}}{\sum_{k=0}^{m-1}t^{k}} =\frac{\sum_{k=0}^{n-1}1}{\sum_{k=0}^{m-1}1} =\frac{n}{m} $$

Answer 2

Binomial expension

$\lim_{x\to1}\frac{x^{1/m}-1}{x^{1/n}-1}$= $ \lim_{h\to 0} \frac{(1+h)^{1/m}-1}{(1+h)^{1/n}-1}$= $ \lim_{h\to 0} \frac{1+\frac{h}{m}+ O(h^2)-1}{1+\frac{h}{n}+O(h^2)-1}$

= $ \lim_{h\to 0} \frac{\frac{h}{m}+ O(h^2)}{\frac{h}{n}+O(h^2)}$ = $ \lim_{h\to 0} \frac{h(\frac{1}{m}+ O(h))}{h (\frac{1}{n}+O(h))}= \frac{n}{m}$

$O(h^2)$= other terms contains $h^2$ and higher powers

Answer 3

A much simpler approach. Let $y=x-1$. Then the expression becomes $\frac{\sqrt[m]{y-1}-1}{ \sqrt[n]{y-1}-1}=\frac{1+\frac{y}{m}+...-1}{1+\frac{y}{n}+...-1}\to \frac{n}{m}$ as $y\to 0$.

Note that $m$ and $n$ do not have to be integers.

Answer 4

Set $x=e^y$, and consider $y \rightarrow 0.$

$f(y):=\dfrac{e^{y/m}-1}{e^{y/n}-1}=$

$\dfrac{1+y/m+O((y/m)^2)-1}{1+y/n+O((y/n)^2) -1}.$

Take the limit.