Calculate the limit $\lim_{x\to 1}\frac{\sqrt[359]{x} - 1}{\sqrt[5]{x} - 1}$ Calculate the limit $$\lim_{x\to 1}\frac{\sqrt[\Large 359]{x} - 1}{\sqrt[\Large 5]{x} - 1}$$
Any hints are appreciated. Cannot use l'Hopital's rule.
 A: Hint: The substitution $x=u^{5\cdot359}$ gets rid of all the radicals, leaving
$$\lim_{u\to1}{u^5-1\over u^{359}-1}$$
Added later (in response to the OP's comment):  You are quite right about the factorizations of $u^5-1$ and $u^{359}-1$, which lead to the limit $5/359$, but I really do mean $x=u^{5\cdot359}$, not $x=u^{1/(5\cdot359)}$.  The radical signs in the original expression mean you're take $x$ to the powers $x^{1/5}$ and $x^{1/359}$, so when you substitute $x=u^n$ to get $x^{1/5}=u^{n/5}$ and $x^{1/359}=u^{n/359}$, you need $n=5\cdot359$, not $n=1/(5\cdot359)$, to get rid of the denominators.
A: You may try below incase you're allowed to use definition of derivative 
$$\begin{align}\lim_{x\to 1}\frac{\sqrt[\Large 359]{x} - 1}{\sqrt[\Large 5]{x} - 1} &=\lim_{x\to 1}\dfrac{\dfrac{\sqrt[\Large 359]{x} - 1}{x-1}}{\dfrac{\sqrt[\Large 5]{x} - 1}{x-1}}\\~\\~\\ &=\dfrac{\lim\limits_{x\to 1}\dfrac{\sqrt[\Large 359]{x} - 1}{x-1}}{\lim\limits_{x\to 1}\dfrac{\sqrt[\Large 5]{x} - 1}{x-1}}\end{align}$$
A: Hint:
$$u^n-1 = (u-1)(1+u+u^2+\dots+u^{n-1})$$
A: It is best to use the standard limit $$\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1}$$ Here we need to use the version dealing with rational $n$ (and this version can be proved using algebraical means). We have then
\begin{align}
L &= \lim_{x \to 1}\frac{\sqrt[359]{x} - 1}{\sqrt[5]{x} - 1}\notag\\
&= \lim_{x \to 1}\frac{\sqrt[359]{x} - 1}{x - 1}\cdot\frac{x - 1}{\sqrt[5]{x} - 1}\notag\\
&= \lim_{x \to 1}\dfrac{\dfrac{\sqrt[359]{x} - 1}{x - 1}}{\dfrac{x - 1}{\sqrt[5]{x} - 1}}\notag\\
&= \dfrac{\dfrac{1}{359}\cdot 1^{1/359 - 1}}{\dfrac{1}{5}\cdot 1^{1/5 - 1}}\notag\\
&= \frac{5}{359}
\end{align}
