$\lim _{ x\to 1 }{ \frac { \sqrt { x } +\sqrt [ 3 ]{ x } -2 }{ x-1 } } $ I have been trying to algebraically solve this limit problem  without using L'Hospital's rule but was whatsoever unsuccessful: 
$$\lim _{ x\to 1 }{ \frac { \sqrt { x } +\sqrt [ 3 ]{ x } -2 }{ x-1 }  }  $$
Thanks in advance!

Let $x = u^6$.
$$\lim _{ x\to 1 }\dfrac { \sqrt { x } +\sqrt [ 3 ]{ x } -2 }{ x-1 } = \lim_{u\to 1}\dfrac{u^3 + u^2 - 1 - 1}{u^6-1}$$
$$
=  \lim_{u\to 1}\dfrac{u^3 - 1}{(u^2)^3-1^3} +  \lim_{u\to 1}\dfrac{u^2 - 1}{(u^2)^3-1^3}
$$
$$
= \lim_{u\to 1}\dfrac{(u - 1)(u^2+u+1)}{(u-1)(u+1)(u^4 + u^2 +1)} + \lim_{u\to 1}\dfrac{(u - 1)(u+1)}{(u-1)(u+1)(u^4 + u^2 +1)} 
$$
$$
= \dfrac{3}{6} + \dfrac{2}{6} = \dfrac{5}{6}
$$
