What is the horizontal asymptote of of #(sqrt(2x^2+1)/(3x-5))# ?

Question

I found that as the function went went to positive infinity, the limit was #sqrt2/3# , and as the function went to negative infinity, the limit was #-sqrt2/3# . So, I am not sure if there is no horizontal asympote since the one sided limits are not equal or if there are two horizontal asymptotes.

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Answer 1 · 2018-03-23T15:06:30

it's #sqrt(2)/3#

Take out a factor of #x^2# from the square root of the numerator. This makes the fraction:
#x(sqrt(2+1/x^2))/(3x-5)#

Then divide top and bottom by x to obtain:

#sqrt(2+1/x^2)/(3-5/x)#

Take the limit as #x# tends to infinity to get the answer