Know your limits! (II)

Without using L'Hopital's rule, we can write the expression as $\lim_{x\rightarrow 1}\frac{1}{\frac{x^2+x-\sqrt{x}-1}{\sin(x^2-1)}}$ $\lim_{x\rightarrow 1}\frac{1}{\frac{x^2-1}{\sin(x^2-1)}+\frac{x-\sqrt{x}}{\sin(x^2-1)}}$ $\lim_{x\rightarrow 1}\frac{1}{\frac{x^2-1}{\sin(x^2-1)}+\frac{\sqrt{x}(\sqrt{x}-1)}{\sin(x^2-1)}}$ $\lim_{x\rightarrow 1}\frac{1}{\frac{1}{\frac{\sin(x^2-1)}{x^2-1}}+\frac{1}{\frac{\sin(x^2-1)}{\sqrt{x}(\sqrt{x}-1)}}}$ now, we know that $\lim_{x\rightarrow 1}\frac{\sin(x^2-1)}{x^2-1}=1,$ the other one needs a little bit of extra work $\lim_{x\rightarrow 1}\frac{1}{\frac{1}{1}+\frac{1}{\frac{\sin(x^2-1)}{\sqrt{x}(\sqrt{x}-1)} \frac{\sqrt{x}+1}{\sqrt{x}+1} }}$ $\lim_{x\rightarrow 1}\frac{1}{1+\frac{1}{\frac{\sin(x^2-1)(\sqrt{x}+1)}{\sqrt{x}(x-1)} \frac{x+1}{x+1}}}$

$\lim_{x\rightarrow 1}\frac{1}{1+\frac{1}{\frac{\sin(x^2-1)(\sqrt{x}+1)(x+1)}{\sqrt{x}(x^2-1)}}}$ we can apply the same limit as before $\lim_{x\rightarrow 1}\frac{1}{1+\frac{1}{\frac{(\sqrt{x}+1)(x+1)}{\sqrt{x}}}}$ now we can take the limit $\frac{1}{1+\frac{1}{4}}=\frac{4}{5},$ so the answer is $4+5=9$

$\begin{aligned} \lim _{ x\to \: 1 } \left( \frac { \sin \left( x^{ 2 }-1 \right) }{ x^{ 2 }+x-\sqrt { x } -1 } \right) & =\lim _{ x\to \: 1 } \left( -\frac { \sin \left( 1-x^{ 2 } \right) }{ x^{ 2 }+x-\sqrt { x } -1 } \right) & \\ & =-\lim _{ x\to \: 1 } \left( \frac { \sin \left( 1-x^{ 2 } \right) }{ x^{ 2 }+x-\sqrt { x } -1 } \right) & \color{grey}{ \lim _{ x\to a } \left[ c\cdot f\left( x \right) \right] =c\cdot \lim _{ x\to a } f\left( x \right) } \\ Applying\, \, \, L'Hopital's\, \, \, Rule: & =-\lim _{ x\to \: 1 } \left( \frac { -2x\cos \left( -x^{ 2 }+1 \right) }{ 2x-\frac { 1 }{ 2\sqrt { x } } +1 } \right) & \\ & =-\lim _{ x\to \: 1 } \left( \frac { -4x^{ \frac { 3 }{ 2 } }\cos \left( 1-x^{ 2 } \right) }{ 4x^{ \frac { 3 }{ 2 } }+2\sqrt { x } -1 } \right) & \\ Plug\, \, \, in\, \, \, values & =-\frac { -4\cdot \: 1^{ \frac { 3 }{ 2 } }\cos \left( 1-1^{ 2 } \right) }{ 4\cdot \: 1^{ \frac { 3 }{ 2 } }+2\cdot \sqrt { 1 } -1 } & \\ \therefore \frac { a }{ b } & =\frac { 4 }{ 5 } \\ \Longrightarrow a+b & =4+5 & \\ & =\boxed{9} & \end{aligned}$

Applying L'Hopital's rule, $\lim_{x\rightarrow1} \frac{\sin (x^2-1)}{x^2+x-\sqrt{x}-1} =\lim_{x\rightarrow1} \frac{2x\cos (x^2-1)}{2x+1-\frac{1}{2\sqrt{x}}}=\frac{2}{2+1-\frac{1}{2}}=\frac{2}{3-\frac{1}{2}}= \frac{4}{6-1}=\frac{4}{5}.$

So a=4 and b=5. $a+b=\boxed{9}$