What the limit!!

Calculus Level 5

lim ⁡ x → 3 ( x − 3 ) ζ ′ ( x 2 − 3 x − 2 ) x 4 − 9 x 3 + 30 x 2 − 44 x + 24 \large{\displaystyle \lim_{x\to 3} \frac { \left( x-3 \right) { \zeta }^{ \prime }\left( { x }^{ 2 }-3x-2 \right) }{ { x }^{ 4 }-9{ x }^{ 3 }+30{ x }^{ 2 }-44x+24 } }

If the value of the above expression is equal to − ζ ( A ) B π C \large{-\frac { \zeta \left( A \right) }{ B{ \pi }^{ C } } }

Find A × B × C A\times B\times C

ζ ′ ( s ) \zeta^{\prime} (s) is first derivative of ζ ( s ) \zeta (s)


The answer is 24.

Tanishq Varshney by Tanishq Varshney , 23, India

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1 solution

Aman Rajput
May 15, 2021

Simplify the denominator as lim ⁡ x → 3 ( x − 3 ) ζ ′ ( x 2 − 3 x − 2 ) ( x − 3 ) ( x − 2 ) 3 \lim_{x\to 3}\frac{(x-3)\zeta'(x^2-3x-2)}{(x-3)(x-2)^3} = lim ⁡ x → 3 ζ ′ ( x 2 − 3 x − 2 ) ( x − 2 ) 3 =\lim_{x\to 3}\frac{\zeta'(x^2-3x-2)}{(x-2)^3} = ζ ′ ( − 2 ) 1 3 = ζ ′ ( − 2 ) =\frac{\zeta'(-2)}{1^3}=\zeta'(-2) Using the property ζ ′ ( − 2 n ) = ( − 1 ) n ζ ( 2 n + 1 ) ( 2 n ) ! 2 2 n + 1 π 2 n \zeta'(-2n)=\frac{(-1)^n\zeta(2n+1)(2n)!}{2^{2n+1}\pi^{2n}} We get answer as − ζ ( 3 ) 4 π 2 \boxed{-\frac{\zeta(3)}{4\pi^2}}

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