What is the limit of #f(x) = -1/(2(lnx)^2)# ?

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Answer 1 · 2017-11-16T01:21:32

#lim_(x->+oo) -1/(2(lnx)^2) =0#
#lim_(x->1) -1/(2(lnx)^2) =-oo#

#f(x) = -1/(2(lnx)^2)#

#lnx# is defined for #x>0 -> f(x)# is defined for #x>0#
Also, #ln(1) =0 -> f(x)# is undefined at #(x=1)#

Hence, the domain of #f(x)# is #(0,+1)uu(+1,+oo)#

Since the question does not specify which limit is sought, I will deal with both possibilities.

#lim_(x->+oo) -1/(2(lnx)^2) -> -1/oo = 0#

Since #lim_(x->1) (lnx)^2 =0# for #1^-# and #1^+#

#lim_(x->1) -1/(2(lnx)^2) -> -1/0 =-oo#

We can deduce these results from the graph of #f(x)# below.

graph{ -1/(2(lnx)^2) [-3.024, 9.463, -3.62, 2.617]}