Find the limit using L'Hopital's rule. Please show the steps.?

Question

lim (ln cot x)^tan x
x tends to 0

Answer 1

= e^ ln (lim (ln cot x)^ tan x)
= e^ (lim ln (ln cot x)^ tan x)
= e^ (lim tan x * ln (ln cot x))

we will concentrate on the limit now

= lim [ ln ( ln cot x) )] / [cot x] ........... this is now an infinity over infinity
= lim 1/(ln cot x) * 1/ cot x * -csc^2 x / -csc^2 x
= lim 1/ [cot x (ln cot x)] ... 1 / infinity ...
= 0

going back to the question
e^0 = 1

lim (ln cot x)^tan x = 1.


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Answer 2

lim x-->0 sin 4x / (x^2+3x) = lim x-->0 4 cos 4x / (2x+3) = lim x-->0 -sixteen sin 4x / 2 = 0 2) x^sin x = e^ ln x^sin x = e^sin x ln x enable y = sin x ln x y = ln x / (a million/sin x ) = 0/0 kind persist with L'Hopital's Rule: lim x-->0 (a million/x) / (-cos x/sin^2 x) lim x-->0 - sin^2 x /x cos x lim x-->0 - 2 sin x cos x / (cos x - x sin x) = 0/(a million-0) = 0/a million =0 y has shrink 0, so e^y has shrink e^0 = a million