$\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!

A scheme sometimes useful is to make the limit at zero...
Let $x=1+t$ so that $x \to 1$ means $t \to 0$.  Then
$$
\frac{\sqrt{x}+\sqrt[3]{x}-2}{x-1} = \frac{\sqrt{1+t}+\sqrt[3]{1+t}-2}{t}
$$
so compute some asymptotics:
$$
\sqrt{1+t} = 1+\frac{1}{2}t+o(t)
\\
\sqrt[3]{1+t} = 1+\frac{1}{3}t+o(t)
\\
\sqrt{1+t}+\sqrt[3]{1+t}-2 = \frac{5}{6}t + o(t)
\\
\frac{\sqrt{1+t}+\sqrt[3]{1+t}-2}{t} = \frac{5}{6}+o(1)
$$
and your limit is $5/6$.
