<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] By the L'Hospital's rule we obtain:  $$\lim_{x\rightarrow1}\frac{x^{\frac{1}{5}}-1}{x^{\frac{1}{6}}-1}=\lim_{x\rightarrow1}\frac{\frac{1}{5}x^{-\frac{4}{5}}}{\frac{1}{6}x^{-\frac{5}{6}}}=\frac{\frac{1}{5}}{\frac{1}{6}}=\frac{6}{5}$$
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