Limited edition

For you who wondered what is the answer

Step 1 : Finding the value of $a$ + $b$ + $c$

$\lim _{ x\xrightarrow [ ]{ } 1 }{ \frac { \sqrt { a{ x }^{ 2 }+bx+c } -3 }{ { (x-1) }^{ 2 } } . } \lim _{ x\xrightarrow [ ]{ } 1 }{ { (x-1) }^{ 2 } } =5.0$

$\sqrt { a+b+c } -3=0$

$a+b+c=9$

Step 2 : Finding another equation

$\lim _{ x\xrightarrow [ ]{ } 1 }{ \frac { \sqrt { a{ x }^{ 2 }+bx+c } -3 }{ { (x-1) }^{ 2 } } . } \lim _{ x\xrightarrow [ ]{ } 1 }{ { (x-1) } } =5.0$

$\lim _{ x\xrightarrow [ ]{ } 1 }{ (\frac { a{ x }^{ 2 }+bx+c-9 }{ { (x-1) } } . } \frac { 1 }{ \sqrt { a{ x }^{ 2 }+bx+c } +3 } )=0$

$\lim _{ x\xrightarrow [ ]{ } 1 }{ (\frac { a{ (x^{ 2 }-1) }+b(x-1) }{ { (x-1) } } . } \frac { 1 }{ \sqrt { a+b+c } +3 } )=0$

$\frac { 1 }{ 6 } \lim _{ x\xrightarrow [ ]{ } 1 }{ (a{ ({ x }+1) }+b) } =0$

$\frac { 2a+b }{ 6 } =0$

$2a=-b$ and it makes $c=9+a$

Step 3 : Finding the value of a

$\lim _{ x\xrightarrow [ ]{ } 1 }{ \frac { \sqrt { a{ x }^{ 2 }+bx+c } -3 }{ { (x-1) }^{ 2 } } } =5$

$\lim _{ x\xrightarrow [ ]{ } 1 }{ (\frac { a{ x }^{ 2 }+bx+c-9 }{ { (x-1) }^{ 2 } } . } \frac { 1 }{ \sqrt { a{ x }^{ 2 }+bx+c } +3 } )=5$

$\lim _{ x\xrightarrow [ ]{ } 1 }{ \frac { a{ x }^{ 2 }-2ax+a }{ { (x-1) }^{ 2 } } . } \frac { 1 }{ \sqrt { a+b+c } +3 } =5$

$\frac { a }{ 6 } =5$

$a = 30$

now we can substitute $a = 30$ to the equation $c=9+a$ we get $c=39$ so the value of $a+c=69$