How do you determine the limit of #5^x/(3^x+2^x)# as x approaches infinity?

1 Answer
Mar 5, 2015

Use l'Hopital's Rule and some algebra.

In the original form, l'Hopital's rule doesn't help, but note that:

#5^x/(3^x+2^x)=(5^x/2^x)/(3^x/2^x+1)=(5/2)^x/((3/2)^x+1)#.

The limit as #xrarroo# is of the form #oo/oo#.

L'Hopital tells us to consider the ratio of the derivatives:

#((5/2)^xln(5/2))/((3/2)^xln(3/2)#.

Note that this is simply #(5/2)^x/(3/2)^x# times the positive constant #ln(5/2)/ln(3/2)#.

Furthermore, #(5/2)^x/(3/2)^x=(5/3)^x#.

So, as #xrarroo#, #5^x/(3^x+2^x)# behaves like #((5/2)^xln(5/2))/((3/2)^xln(3/2)# which in turn behaves like #k(5/3)^x# with #k>0#. Which increases without bound (goes to #oo#) as #xrarroo#.

#lim_(xrarroo)5^x/(3^x+2^x)=oo#.

Bonus: #lim_(xrarr-oo)(5/2)^x/((3/2)^x+1)=0/(0+1)# so #lim_(xrarr-oo)5^x/(3^x+2^x)=0#