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Calculus Level 5

$\lim _{x\rightarrow2}{\left(\frac { \pi \ln { \left| 2\sin { \pi x } \right| } }{ 2\sinh ^{ 2 }{ \pi x } } -\frac { \pi \ln { \left| 16-{ x }^{ 4 } \right| } }{ 2\sinh ^{ 2 }{ 2\pi } }\right) } = \frac { A\pi \ln { B\pi } -\ln { C } }{ D\sinh ^{ E }{ F\pi } }$

If the equation above holds true for positive integers $A$ , $B$ , $C$ , $D$ , $E$ , and $F$ , find $A+B+C+D+E+F$ .

The answer is 24.

1 solution

Chew-Seong Cheong
Nov 30, 2017

$\begin{aligned} L & = \lim_{x \to 2} \left(\frac {\pi \ln |2\sin \pi x|}{2 \sinh^2 \pi x} - \frac {\pi \ln |16- x^4|}{2 \sinh^2 \pi x} \right) \\ & = \lim_{x \to 2} \frac \pi {2\sinh^2 \pi x} \cdot \ln \left|\frac {2\sin \pi x}{16- x^4} \right| & \small \color{#3D99F6} \text{A 0/0 cases, L'Hôpital's rule applies.} \\ & = \lim_{x \to 2} \frac \pi {2\sinh^2 \pi x} \cdot \ln \left|\frac {2 \pi \cos \pi x}{-4 x^3} \right| & \small \color{#3D99F6} \text{Differentiate us and down w.r.t. }x. \\ & = \frac \pi {2\sinh^2 2 \pi} \cdot \ln \left|\frac {2 \pi}{-32} \right| \\ & = \frac {\pi \ln \pi - \ln 16 }{2\sinh^2 2 \pi} \end{aligned}$

$\implies A+B+C+D+E+F = 1+1+16+2+2+2 = \boxed{24}$

Who's up to the challenge? 73

The answer is 24.

1 solution

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