Lim x___0 sinhx -sinx/x^3 ? using L’ Hopital’s rule

1 Answer
Ultrilliam
May 25, 2018

#= 1/3#

Explanation:

This is in #0/0# indeterminate form throughout, allowing multiple L'Hopital application:

#lim_(x to 0) (sinhx -sinx)/x^3#

#= lim_(x to 0) (coshx - cosx)/(3 x^2)#

#= lim_(x to 0) (sinhx + sinx)/(6 x)#

#= lim_(x to 0) (coshx + cosx)/(6 ) #

#= (lim_(x to 0) coshx + lim_(x to 0) cosx)/(6 ) = (1 + 1)/6#

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