<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] Correct me if wrong:
Let $y^{30}:=x$.
Then: $ x^{1/5} = y^6 $; $x^{1/6} =y^5$.
Numerator: 
$n(y)=(y^6-1) =$
$(y-1)(y^5+y^4+y^3+y^2+y+1);$
Denominator: 
$d(y) =(y^5-1)= $
$(y-1)(y^4+y^3+y^2+y+1).$
$\lim_{y \rightarrow 1}\dfrac{n(y)}{d(y)}=$
$\lim_{y \rightarrow 1}\dfrac{y^5 +y^4+y^3+y^2+y+1}{y^4+y^3+y^2+y+1}=6/5.$
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