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

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Answer 1 · 2016-07-31T13:56:20

1

#L = lim_(x to 0) x^(3 sin x)#

#ln L = lim_(x to 0) ln x^(3 sin x) = lim_(x to 0) 3 sin x ln x#

#= 3 lim_(x to 0) ln x / (csc x)# which is now #oo/oo# indeterminate so we can use LHopital's Rule

#= 3 lim_(x to 0) (1/ x) / (-csc x cot x)#

#= - 3 lim_(x to 0) (sin x tan x) / (x)# which is #0/0# indterminate so again LHopital

#= - 3 lim_(x to 0) (cos x tan x + sin x sec^2 x) / (1)#

#= - 3 lim_(x to 0) sin x + tan x sec x = 0#

so # lim_(x to 0) ln L = 0#

#implies lim_(x to 0) L = 1#