Question 157930
Looking at the numerator {{{8x}}}, we can see that the degree is {{{1}}} since the highest exponent of the numerator is {{{1}}}. For the denominator {{{2x^2+1}}}, we can see that the degree is {{{2}}} since the highest exponent of the denominator is {{{2}}}.


Since the degree of the numerator (which is {{{1}}}) is less than the degree of the denominator (which is {{{2}}}), the horizontal asymptote is always {{{y=0}}}


So the horizontal asymptote is {{{y=0}}}



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Notice if we graph {{{y=(8x)/(2x^2+1)}}}, we can visually verify our answer:


{{{drawing(500,500,-10,10,-10,10,
graph(500,500,-10,10,-10,10,(8x)/(2x^2+1)),
green(line(-20,0,20,0))
)}}} Graph of {{{y=(8x)/(2x^2+1))}}} with the horizontal asymptote {{{y=0}}} (green line)