How do you find the limit #lnx/x# as #x->oo#?

2 Answers
Nov 5, 2017

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

Explanation:

If we evaluate the limit of the numerator and denominator separately we'll find that:

#*# As #ln(x)# goes to #oo# as #x# goes to #oo#: #ln(oo)=oo#

#*# #x# goes to #oo#

Therefore we have a ratio of two infinities #oo/oo# meaning that we will have to apply L'Hospital's Rule.

#lim_(x->oo)lnx/x=lim_(x->oo)(d/dx(lnx))/(d/dx(x))=lim_(x->oo)(1/x)/1=lim_(x->oo)1/x=0#

The limit approaches #0# because #1# divided over something approaching #oo# becomes closer and closer to #0#

For example, consider:

#1/10=0.1#

#1/100=0.01#

#1/10000=0.0001#

We can see that as the denominator gets larger and larger, approaching #oo#, the value gets smaller and smaller and more closer to #0#.

Nov 5, 2017

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

Explanation:

The question is to find the value of #lnx/x# where #x to oo#

If we let #x=oo# then #lnx/x=(oo)/(oo)##=# Undefined

So now we can apply L'Hospital's rule by differentiating the numerator and denominator individually.

So, #d/dxlnx=1/x# and #d/dxx=1#

#lim_(x -> oo)(1/x)/1#

#lim_(x -> oo)1/x#

Now we let #x=oo#

Therefore #1/(oo)=0# because #oo# is a very large number and #1# divided by a very large number always approaches #0# and it is very close to #0# but never quite gets there.