How do you find the limit of # ln ( 3x + 5e^x )/ ln ( 7x + 3e^{2x})# as x approaches infinity?

1 Answer
Cesareo R.
Jul 3, 2016

#lim_{x->oo}log_e ( 3x + 5e^x )/ log_e (7x+3e^{2x}) =1/2#

Explanation:

#log_e ( 3x + 5e^x )/ log_e ( 7x + 3e^{2x})#

For large #x# follows that
#log_e ( 3x + 5e^x ) approx log_e (5e^x ) #
#log_e ( 7x + 3e^{2x}) approx log_e (3e^{2x})#

so
#lim_{x->oo}log_e ( 3x + 5e^x )/ log_e (7x+3e^{2x}) = lim_{x->oo}log_e (5e^x )/ log_e (3e^{2x})#

but

# lim_{x->oo}log_e (5e^x )/ log_e (3e^{2x}) = lim_{x->oo}(log_e 5 + x)/(log_3 3+2x)= 1/2#

Related questions
See all questions in Determining Limits Algebraically
Impact of this question
4863 views around the world
You can reuse this answer
Creative Commons License