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:
$$       \lim_{x \to 1}  \frac{   (x^\frac{1}{5} - 1) (x^\frac{1}{6}  + 1)        }{                  (x^\frac{1}{6} - 1) (x^\frac{1}{6}  + 1)        }     $$
$$\lim_{x \to 1}  \frac{   (x^\frac{1}{5} - 1) (x^\frac{1}{6}  + 1)        }{                  (x^\frac{2}{6} - 1)   } $$
 A: 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 $\frac{6}{5}$.
A: By L'Hopital's rule:\begin{align}\lim_{x\to1}\frac{x^\frac15-1}{x^\frac16-1}&=\lim_{x\to1}\frac{\frac15x^{-\frac45}}{\frac16x^{-\frac56}}\\&=\frac65.\end{align}
A: By the L'Hospital's rule we obtain:  $$\lim_{x\rightarrow1}\frac{x^{\frac{1}{5}}-1}{x^{\frac{1}{6}}-1}=\lim_{x\rightarrow1}\frac{\frac{1}{5}x^{-\frac{4}{5}}}{\frac{1}{6}x^{-\frac{5}{6}}}=\frac{\frac{1}{5}}{\frac{1}{6}}=\frac{6}{5}$$
A: Let $x=1+y$ with $y\to0$ and use
$$(1+y)^a=1+ay+o(y)$$
$$\frac{\left(x^\frac{1}{5}\right)-1}{ \left( x^\frac{1}{6}\right) -1}=\frac{1+\frac{1}{5}y-1+o(y)}{1+ \frac{1}{6}y -1+o(y)}=\frac{\frac{1}{5}y+o(y)}{\frac{1}{6}y+o(y)}=\frac{\frac{1}{5}+o(1)}{\frac{1}{6}+o(1)}\to\frac65$$
A: 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)= $
$(y-1)(y^4+y^3+y^2+y+1).$
$\lim_{y \rightarrow 1}\dfrac{n(y)}{d(y)}=$
$\lim_{y \rightarrow 1}\dfrac{y^5 +y^4+y^3+y^2+y+1}{y^4+y^3+y^2+y+1}=6/5.$
A: Hint:
As lcm$(5,6)=30,$ choose $x^{1/30}=y\implies x^{1/5}=y^6, x^{1/6}=?$
A: 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 $x\to 1$
A: 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! :)
A: By the generalized binomial theorem,
$$(1+t)^\alpha-1=\alpha t+\frac{\alpha(\alpha-1)}2t^2+\frac{\alpha(\alpha-1)(\alpha-2)}{3!}t^3+\cdots$$ and higher order terms.
Hence your limit is essentially
$$\lim_{t\to0}\frac{\dfrac t5}{\dfrac t6}.$$
