How do you find the limit of #(tan 3x)^x# as x approaches 0?

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Answer 1 · 2016-07-13T13:30:07

1

#lim_{x to 0} (tan 3x)^x#

#=lim_{x to 0} e^{ ln ((tan 3x)^x)}#

#= exp ( lim_{x to 0} x ln ((tan 3x)) )# as e^x is continuous function we
can lift it outside the limit

#= exp ( lim_{x to 0} ( ln ((tan 3x)) )/(1/x))# which is indeterminate so we can use LHospital's rule

#= exp ( lim_{x to 0} ( (3sec^2 3x) /((tan 3x)) )/(1/x^2))#

#= exp ( lim_{x to 0} 1/cos^2 (3x) * (3x^2) /(tan 3x) )# and we can lift the cos term out as continuous and apply L'Hopital again

#= exp (1/cos^2 (3x) * lim_{x to 0} (6x) /(3 sec^2 3x) )#

#= exp (1/cos^2 (3x) * lim_{x to 0} (6x) * (3 cos^2 3x) )#

#= exp (1 * 0 * 3) = 1#