Know your limits! (III)

Applying L'Hopital's rule, $\lim_{x\rightarrow1} \frac{\sqrt{2x-1}-\sqrt{x}}{x^2+x-2}= \lim_{x\rightarrow1} \frac{\frac{2}{2\sqrt{2x-1}}- \frac{1}{2\sqrt{x}}}{2x+1} = \frac{\frac{2}{2}- \frac{1}{2}}{2+1}=\frac{\frac{1}{2}}{3}=\frac{1}{6}.$ a=1, b=6. $a+b=\boxed{7}$

Apply L'hopital's rule.

$\begin{aligned} \lim _{ x\to \: 1 } \left( \frac { \sqrt { 2x-1 } -\sqrt { x } }{ x^{ 2 }+x-2 } \right) & =\lim _{ x\to \: 1 } \left( \frac { \frac { 1 }{ \sqrt { 2x-1 } } -\frac { 1 }{ 2\sqrt { x } } }{ 2x+1 } \right) \\ & =\lim _{ x\to \: 1 } \left( \frac { 2\sqrt { x } -\sqrt { 2x-1 } }{ 2\sqrt { x } \sqrt { 2x-1 } \left( 2x+1 \right) } \right) \\ Plug\, \, \, in\, \, \, values & :\, \, \, \frac { 2\cdot \sqrt { 1 } -\sqrt { 2\cdot \: 1-1 } }{ 2\cdot \sqrt { 1 } \sqrt { 2\cdot \: 1-1 } \left( 2\cdot \: 1+1 \right) } \\ \therefore \frac { a }{ b } & =\frac { 1 }{ 6 } \\ \Longrightarrow a+b & =1+6 \\ & =\boxed{7} \end{aligned}$