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:
$$\lim_{ x \to 1}\frac{\sqrt[n]{(u+1)^n-1}}{\sqrt[n]{n(u+1)}-\sqrt[n]{n}-\sqrt[n]{n(u+1)-n}}$$
We can simplify the term of interest and rationalize terms to obtain
$$\begin{align} \frac{\sqrt[n]{x^n-1}}{\sqrt[n]{nx}-\sqrt[n]{n}-\sqrt[n]{nx-n}}&=\frac{\sqrt[n]{x^n-1}}{\sqrt[n]{n}\,(\,\sqrt[n]{x}\,-1\,-\,\sqrt[n]{x-1}\,)}\\\\ &=\frac{\sqrt[n]{x^{n-1}+x^{n-2}+\cdots +1}\,\,\sqrt[n]{x-1}}{\sqrt[n]{n}\,(\,\sqrt[n]{x}\,-1\,-\,\sqrt[n]{x-1}\,)}\\\\ &=\left(\frac{\sqrt[n]{x^{n-1}+x^{n-2}+\cdots +1}}{\sqrt[n]{n}}\right)\left(\frac{\sqrt[n]{x-1}}{\sqrt[n]{x}\,-1\,-\,\sqrt[n]{x-1}}\right)\\\\ &=\left(\frac{\sqrt[n]{x^{n-1}+x^{n-2}+\cdots +1}}{\sqrt[n]{n}}\right)\left(\frac{\sqrt[n]{x-1}}{\frac{x-1}{\sqrt[n]{x^{n-1}+x^{n-2}+\cdots +1}}-\sqrt[n]{x-1}}\right)\\\\ &=\left(\frac{\sqrt[n]{x^{n-1}+x^{n-2}+\cdots +1}}{\sqrt[n]{n}}\right)\left(\frac{1}{\frac{\sqrt[n]{(x-1)^{n-1}}}{\sqrt[n]{x^{n-1}+x^{n-2}+\cdots +1}}-1}\right)\\\\ &\to \left(\frac{\sqrt[n]{n}}{\sqrt[n]{n}}\right)\left(\frac{1}{\frac{0}{\sqrt[n]{n}}-1}\right)=-1 \end{align}$$
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}