What is the horizontal asymptote of #y=((2x+3)(5x-2))/(7x^2-3)# ?

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What is the horizontal asymptote of #y=((2x+3)(5x-2))/(7x^2-3)# ?

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Answer 1 · 2014-09-11T00:21:22

Its horizontal asymptote is #y=10/7#.

By taking the limits at infinity,
#lim_{x to infty}{(2x+3)(5x-2)}/{7x^2-3}#
by divide the numerator and the denominator by #x^2#,
#=lim_{x to +infty}{(2+3/x)(5-2/x)}/{7-3/x^2}={(2+0)(5-0)}/{7-0}=10/7#

Similarly, you can find
#lim_{x to -infty}{(2x+3)(5x-2)}/{7x^2-3}=10/7#

Hence, there is only one horizontal asymptote #y=10/7#.

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What is the horizontal asymptote of #y=((2x+3)(5x-2))/(7x^2-3)# ?

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