<s>[INST] 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:
$$       \lim_{x \to 1}  \frac{   (x^\frac{1}{5} - 1) (x^\frac{1}{6}  + 1)        }{                  (x^\frac{1}{6} - 1) (x^\frac{1}{6}  + 1)        }     $$
$$\lim_{x \to 1}  \frac{   (x^\frac{1}{5} - 1) (x^\frac{1}{6}  + 1)        }{                  (x^\frac{2}{6} - 1)   } $$
 [/INST] Here's another solution depending on what you're allowed to use: substitute $u=x^{\frac{1}{6}}$.
Then we get 
$$\lim_{u\rightarrow 1}\frac{u^{\frac{6}{5}}-1}{u-1}$$
and we can recognise this as the derivative of $u\mapsto u^{\frac{6}{5}}$ at 1, which is of course $\frac{6}{5}$.
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