<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] Just to avoid L' Hospital's rule, consider the following:
$$\frac{x^{1/5}-1}{x^{1/6}-1}=\frac{x^{6/30}-1}{x^{5/30}-1}=\frac{(x^{1/30})^6-1}{(x^{1/30})^5-1}=\frac{(x^{1/30}-1)(x^{5/30}+x^{4/30}+\dots+1)}{(x^{1/30}-1)(x^{4/30}+x^{3/30}+\dots+1)}$$
So:
$$\lim_{x\to1}\frac{x^{1/5}-1}{x^{1/6}-1}=\lim_{x\to1}\frac{(x^{1/30}-1)(x^{5/30}+x^{4/30}+\dots+1)}{(x^{1/30}-1)(x^{4/30}+x^{3/30}+\dots+1)}=\lim_{x\to1}\frac{x^{5/30}+x^{4/30}+\dots+1}{x^{4/30}+x^{3/30}+\dots+1}=\frac{6}{5}$$
Hope this provided an alternative! :)
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