$\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!

Or without using L'Hospital, you can do this:
$$
\lim _{ x\rightarrow 1 }{ \frac { \sqrt { x } +\sqrt [ 3 ]{ x } -2 }{ x-1 }  } \quad =\quad \lim _{ x\rightarrow 1 }{ \frac { \sqrt { x } -1+\sqrt [ 3 ]{ x } -1 }{ x-1 }  } \\ \qquad \qquad \qquad \qquad \qquad \quad =\quad \quad \lim _{ x\rightarrow 1 }{ \frac { \sqrt { x } -1 }{ x-1 }  } +\frac { \sqrt [ 3 ]{ x } -1 }{ x-1 } \\ \qquad \qquad \qquad \qquad \qquad \quad =\quad \lim _{ x\rightarrow 1 }{ \frac { \sqrt { x } -1 }{ \left( \sqrt { x } -1 \right) \left( \sqrt { x } +1 \right)  }  } +\frac { \sqrt [ 3 ]{ x } -1 }{ \left( \sqrt [ 3 ]{ x } -1 \right) \left( \sqrt [ 3 ]{ x } +{ \left( \sqrt [ 3 ]{ x }  \right)  }^{ 2 }+1 \right)  } \\ \qquad \qquad \qquad \qquad \qquad \quad =\quad \quad \lim _{ x\rightarrow 1 }{ \frac { 1 }{ \sqrt { x } +1 }  } +\frac { 1 }{ \sqrt [ 3 ]{ x } +{ \left( \sqrt [ 3 ]{ x }  \right)  }^{ 2 }+1 } \\ \qquad \qquad \qquad \qquad \qquad \quad =\quad \frac { 1 }{ 2 } +\frac { 1 }{ 3 } \\ \qquad \qquad \qquad \qquad \qquad \quad =\quad \frac { 5 }{ 6 }  
$$
Identities used:


*

*$a^2-b^2 =(a-b)(a+b) $

*$a^3-b^3 = (a-b)(a^2+ab+b^2)$

