How do you find the limit of #(2^x)/(3^x-2^x)# as #x->oo#?

1 Answer
VNVDVI
Mar 29, 2018

#0#

Explanation:

Plugging in #oo# right away yields

#lim_(x->oo)2^x/(3^x-2^x)=oo/(oo-oo),# an indeterminate form that doesn't really tell us anything, so further simplification is needed.

Note that #2^x# appears in the numerator and denominator. We can then divide the entire limit, numerator and denominator, by #2^x# and have a useful simplification.

#lim_(x->oo)(2^x/2^x)/((3^x/2^x)-(2^x/2^x))=lim_(x->oo)1/((3/2)^x-1)=1/(lim_(x->oo)(3/2)^x-1#

Recall that #lim_(x->oo)r^x=oo# if #|r|>1, 0# if #|r|<1#. Here, #|r|=3/2>1#, so the limit will approach infinity.

#1/(lim_(x->oo)(3/2)^x-1)=1/oo=0#

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