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:
$$\lim_{ x \to 1}\frac{\sqrt[n]{(u+1)^n-1}}{\sqrt[n]{n(u+1)}-\sqrt[n]{n}-\sqrt[n]{n(u+1)-n}}$$

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