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Answer 1 · 2016-12-21T17:19:32

#lim_(xrarroo)e^(3x)/(5x^2) = oo#. Because #e^x# grows faster that any positive power of #x#.

The quick explanation is that #lim_(xrarroo)e^x/x^p = oo# for any positive #p#. (Indeed, for any real #p#.)

My longer explanation uses l'Hospitals's Rule

#lim_(xrarroo)e^(3x)/(5x^2)# has indeterminate initial form #oo/oo#.

Apply l'Hospital and consider

#lim_(xrarroo)(3e^(3x))/(10x)# which also has indeterminate initial form #oo/oo#.

Apply l'Hospital again to get

#lim_(xrarroo)(9e^(3x))/(10)# which is #oo#.

Therefore the original limit is #oo#.