Find the limit : $\lim_{ x \to 1}\frac{\sqrt[n]{x^n-1}}{\sqrt[n]{nx}-\sqrt[n]{n}-\sqrt[n]{nx-n}}$ Find the limit: Without the use of the L'Hôspital's Rule
$$\lim_{ x \to 1}\frac{\sqrt[n]{x^n-1}}{\sqrt[n]{nx}-\sqrt[n]{n}-\sqrt[n]{nx-n}}$$
My try:
$u=x-1$
Now:
$$\lim_{ x \to 1}\frac{\sqrt[n]{(u+1)^n-1}}{\sqrt[n]{n(u+1)}-\sqrt[n]{n}-\sqrt[n]{n(u+1)-n}}$$

Let us assume $n > 1$. We must have $x \to 1^{+}$ to ensure that the roots are well defined for all $n > 1$. We can proceed as follows
\begin{align}
L &= \lim_{x \to 1^{+}}\frac{\sqrt[n]{x^{n} - 1}}{\sqrt[n]{nx} - \sqrt[n]{n} - \sqrt[n]{nx - n}}\notag\\
&= \frac{1}{\sqrt[n]{n}}\lim_{x \to 1^{+}}\frac{\sqrt[n]{x^{n} - 1}}{\sqrt[n]{x} - 1 - \sqrt[n]{x - 1}}
&= \frac{1}{\sqrt[n]{n}}\lim_{x \to 1^{+}}\dfrac{\sqrt[n]{\dfrac{x^{n} - 1}{x - 1}}}{\dfrac{\sqrt[n]{x} - 1}{\sqrt[n]{x - 1}} - 1}\notag\\
&= \lim_{x \to 1^{+}}\dfrac{1}{\dfrac{\sqrt[n]{x} - 1}{x - 1}\cdot (x - 1)^{1 - 1/n} - 1}\notag\\
&= \dfrac{1}{\dfrac{1}{n}\cdot 0 - 1}\notag\\
&= -1\notag
\end{align}
