$\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!

Using de l'hospital rule you get
$$
\lim_{x \to 1} \frac{x^{\frac{1}{2}}+x^{\frac{1}{3}}-2}{x-1} = \left[\frac{0}{0}\right]=\lim_{x \to 1} \frac{\frac{1}{2}x^{-\frac{1}{2}}+\frac{1}{3}x^{-\frac{2}{3}}}{1} = \frac{5}{6}.
$$
