How do get the following
$\displaystyle \displaystyle\lim_{x\to 1}\frac{x^\frac{1}{7}-1}{x^\frac{1}{5}-1}$
Without l'Hopital's Rule....
One way to appraoch it is to do some fancy factoring.
$\displaystyle \lim_{x\rightarrow{1}}\frac{x^{\frac{1}{7}}-1}{x^{\frac{1}{5}}-1}$
=$\displaystyle \lim_{x\rightarrow{1}}\frac{x^{\frac{4}{35}}+x^{\f rac{3}{35}}+x^{\frac{2}{35}}+x^{\frac{1}{35}}+1}{x ^{\frac{6}{35}}+x^{\frac{4}{35}}+x^{\frac{3}{35}}+ x^{\frac{2}{35}}+x^{\frac{1}{7}}+x^{\frac{1}{35}}+ 1}=\boxed{\frac{5}{7}}$
Substitute $\displaystyle u^{35}=x,$ the limit becomes to
$\displaystyle \lim_{u\to1}\frac{u^5-1}{u^7-1}.$
The factor $\displaystyle u-1$ is bothering the top & bottom, so we pull it out:
$\displaystyle u^5-1=(u-1)(u^4+u^3+u^2+u+1),$ $\displaystyle u^7-1=(u-1)(u^6+u^5+u^4+u^3+u^2+u+1),$ so
$\displaystyle \lim_{u\to1}\frac{u^5-1}{u^7-1}=\lim_{u\to1}\frac{u^4+u^3+u^2+u+1}{u^6+u^5+u^4+ u^3+u^2+u+1},$
and the conclusion follows $\displaystyle \blacksquare$