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OpenStudy (anonymous):

for what values of k is lim x---> infinity sinhkx/cosh2x finite

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OpenStudy (anonymous):

use the defintion of sinh and cosh: \[\sinh(x) = {1/2} (e^{x}-e^{-x})\] \[\cosh(x) = {1/2} (e^{x}+e^{-x})\] \[\lim_{x \rightarrow \infty} (e^{kx}-e^{-kx})/(e^{2x}+e^{-2x})\]

OpenStudy (anonymous):

is that the final answer

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