$\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: One may recall that for any differentiable function $f$ around $a$, we have

$$
\frac{f(x)-f(a)}{x-a} \to f'(a).
$$ 

You may easily apply it here with 
$$
f(x)=x^{1/2}+x^{1/3}, \quad a=1.
$$
A: 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) $

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

A: Using de l'hospital rule you get
$$
\lim_{x \to 1} \frac{x^{\frac{1}{2}}+x^{\frac{1}{3}}-2}{x-1} = \left[\frac{0}{0}\right]=\lim_{x \to 1} \frac{\frac{1}{2}x^{-\frac{1}{2}}+\frac{1}{3}x^{-\frac{2}{3}}}{1} = \frac{5}{6}.
$$
A: Let $x = u^6$.
$$\lim _{ x\to 1 }\dfrac { \sqrt { x } +\sqrt [ 3 ]{ x } -2 }{ x-1 } = \lim_{u\to 1}\dfrac{u^3 + u^2 - 1 - 1}{u^6-1}$$
$$
=  \lim_{u\to 1}\dfrac{u^3 - 1}{(u^2)^3-1^3} +  \lim_{u\to 1}\dfrac{u^2 - 1}{(u^2)^3-1^3}
$$
$$
= \lim_{u\to 1}\dfrac{(u - 1)(u^2+u+1)}{(u-1)(u+1)(u^4 + u^2 +1)} + \lim_{u\to 1}\dfrac{(u - 1)(u+1)}{(u-1)(u+1)(u^4 + u^2 +1)} 
$$
$$
= \dfrac{3}{6} + \dfrac{2}{6} = \dfrac{5}{6}
$$
A: 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 $5/6$.
