Question

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

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.

Answer 1

it's `sqrt(2)/3`

Explanation:

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