How do you solve this ? #lim_(x->0^+)(1+sin(4x))^cot(x)#

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Answer 1 · 2017-08-08T22:03:21

#lim_{x\to0^+} (1+sin(4x))^(cot(x)) = e^4#.

#y=(1+sin(4x))^(cot(x))#
#ln(y)=cot(x)ln(1+sin(4x)#,
#ln(y)=ln(1+sin(4x))/(tan(x))#.

Then #lim_{x\to0^+} ln(y)# is in the indeterminate form #0/0#.

By L'Hopitals rule, if #f(a)=g(a)=0# then #lim_{x\toa} (f(a))/(g(a)) = lim_{x\toa} (f'(a))/(g'(a))#.

Then,

#lim_{x\to0^+} ln(y) = lim_{x\to0^+} ((4cos(4x))/(1+sin(4x)))/(sec^2(x))#,
#lim_{x\to0^+} ln(y) = 4#.

Then taking an inverse logarithm,

#lim_{x\to0^+} y = e^4#.