Know your limits! (IV)

Applying L'Hopital's rule, $\lim_{x\rightarrow1} \frac{x^{1/2}-(2-x)^{1/3}}{x^2+x-2} =\lim_{x\rightarrow1} \frac{\frac{1}{2}x^{-1/2}-(-1)\frac{1}{3}(2-x)^{-2/3}}{2x+1} = \lim_{x\rightarrow1} \frac{\frac{1}{2}x^{-1/2}+\frac{1}{3}(2-x)^{-2/3}}{2x+1}$
$=\frac{\frac{1}{2}+\frac{1}{3}}{2+1}= \frac{\frac{5}{6}}{3}=\frac{5}{18}$

So a=5 and b=18. $a+b =\boxed{23}$

$\begin{aligned} \lim _{ x\to 1 } \frac { x^{ 1/2 }-(2-x)^{ 1/3 } }{ x^{ 2 }+x-2 } & =\lim _{ x\to \: 1 } \left( \frac { \sqrt { x } -\sqrt [ 3 ]{ 2-x } }{ x^{ 2 }+x-2 } \right) & \\ Applying\quad L'Hopital's\quad Rule & =\lim _{ x\to \: 1 } \left( \frac { \frac { 1 }{ 2\sqrt { x } } +\frac { 1 }{ 3\left( -x+2 \right) ^{ \frac { 2 }{ 3 } } } }{ 2x+1 } \right) & \\ & =\lim _{ x\to \: 1 } \left( \frac { 3\left( 2-x \right) ^{ \frac { 2 }{ 3 } }+2\sqrt { x } }{ 6\left( 2-x \right) ^{ \frac { 2 }{ 3 } }\sqrt { x } \left( 2x+1 \right) } \right) & \\ Plug\quad in\quad Values & =\frac { 3\left( 2-1 \right) ^{ \frac { 2 }{ 3 } }+2\cdot \sqrt { 1 } }{ 6\left( 2-1 \right) ^{ \frac { 2 }{ 3 } }\sqrt { 1 } \left( 2\cdot \: 1+1 \right) } & \\ \therefore \frac { a }{ b } & =\frac { 5 }{ 18 } & \\ \Longrightarrow a+b & =5+18 & \\ & =23 & \end{aligned}$

nice problem