A power difference

let x=y^2 so as x--->1 y--->1 the problem becomes lim(y^2+y-2)/(y^4+y-2)=lim(y-1)(y+2)/(y-1)(y^3+y^2+y+2) =lim(y+2)/(y^3+y^2+y+2)=3/5=0.6

Applying L'Hopital's rule, $\large \lim_{x\to1 } \dfrac{x + \sqrt x - 2}{x^2 + \sqrt x - 2 } = \large \lim_{x\to1 } \dfrac{1 + \dfrac{1}{2\sqrt x}}{2x + \dfrac{1}{2\sqrt x }}$

$=\dfrac{1+\dfrac{1}{2}}{2+\dfrac{1}{2}} = \dfrac{2+1}{4+1} =\dfrac{3}{5}=\boxed{0.6}$

Factor $\frac{x+\sqrt{x}-2}{x^2+\sqrt{x}-2}$

$\begin{aligned} \lim _{ x\to 1 } \left( \frac { x+\sqrt { x } -2 }{ x^{ 2 }+\sqrt { x } -2 } \right) & =\lim _{ x\to 1 } \left( \frac { \left( \sqrt { x } -1 \right) \left( \sqrt { x } +2 \right) }{ \left( \sqrt { x } -1 \right) \left( x^{ \frac { 3 }{ 2 } }+x+\sqrt { x } +2 \right) } \right) \\ & =\lim _{ x\to \: 1 } \left( \frac { \left( \sqrt { x } -1 \right) \left( \sqrt { x } +2 \right) }{ \left( \sqrt { x } -1 \right) \left( x^{ \frac { 3 }{ 2 } }+x+\sqrt { x } +2 \right) } \right) \\ & =\lim _{ x\to \: 1 } \left( \frac { \sqrt { x } +2 }{ x^{ \frac { 3 }{ 2 } }+x+\sqrt { x } +2 } \right) \\ Plug\, \, \, in\, \, \, values\, \, & :\, \, \frac { \sqrt { 1 } +2 }{ 1^{ \frac { 3 }{ 2 } }+1+\sqrt { 1 } +2 } \\ \therefore \lim _{ x\to 1 } \left( \frac { x+\sqrt { x } -2 }{ x^{ 2 }+\sqrt { x } -2 } \right) & =\frac { 3 }{ 5 } \end{aligned}$