Computing $\lim_{x \to 1}\frac{x^\frac{1}{5}-1}{x^\frac{1}{6} -1}$ I cannot figure out how to
get around the zero numerator and denominator in order to compute the limit below:
$$\lim_{x \to 1}\frac{\left(x^\frac{1}{5}\right)-1}{ \left( x^\frac{1}{6}\right) -1}$$
I tried:
$$       \lim_{x \to 1}  \frac{   (x^\frac{1}{5} - 1) (x^\frac{1}{6}  + 1)        }{                  (x^\frac{1}{6} - 1) (x^\frac{1}{6}  + 1)        }     $$
$$\lim_{x \to 1}  \frac{   (x^\frac{1}{5} - 1) (x^\frac{1}{6}  + 1)        }{                  (x^\frac{2}{6} - 1)   } $$

Take $u=x^{\frac{1}{5}}$ and $v=x^{\frac{1}{6}}$.
$\displaystyle u-1=\frac{u^5-1}{u^4+u^3+u^2+u+1}=\frac{x-1}{u^4+u^3+u^2+u+1}$
$\displaystyle v-1=\frac{v^6-1}{v^5+v^4+v^3+v^2+v+1}=\frac{x-1}{v^5+v^4+v^3+v^2+v+1}$
$\displaystyle \frac{u-1}{v-1}=\frac{v^5+v^4+v^3+v^2+v+1}{u^4+u^3+u^2+u+1}\to\frac{6}{5}$ as $x\to 1$
