How do you find the limit of #(cot(x)) / (ln(x))# as x approaches 0?

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Answer 1 · 2016-09-21T11:30:43

#-oo#

#cot(x)/ln(x)=cos(x)/(sin(x)ln(x)) #

but

#sin(x)ln(x) = ln(x^sin(x))# and

now, making #x = 1 + delta#

#x^sin(x) = (1+delta)^sin(1+delta)#

and

#(1+delta)^sin(1+delta) = 1 + sin(1+delta)delta+(sin(1+delta)(sin(1+delta)-1))/(2!)delta^2+ cdots#

then

#lim_(x->0) ln(x^sin(x))=ln(lim_(x->0) x^sin(x)) = #
#=ln(lim_(delta->-1)(1+delta)^sin(1+delta) ) = ln(1^-)=0^-#

so

#lim_(x->0)cot(x)/ln(x) = -oo#