What are the vertical and horizontal asymptotes of this function? 4x / x^2+1?

Question

please give detail

Answer 1

f(x) = 4x / (x^2 + 1)

To solve for the vertical asymptotes, all you have to do is equate the denominator to 0; the result of doing so will tell us our asymptote. In this particular case, x^2 + 1 = 0 has no real solutions, so there are no vertical asymptotes.

To solve for the horizontal asymptotes, take the limit as x approaches infinity and x approaches negative infinity.

lim ( 4x / (x^2 + 1) )
x -> infinity

Divide each term by x^2, and we get

lim ( [4/x] / (1 + 1/x^2) )
x -> infinity

Which we can now evaluate directly.

0/(1 + 0) = 0

Therefore, the limit is 0, which means we have a horizontal asymptotes at y = 0. Similarly for negative infinity, we should get 0 as well.

Answer 2

You can do it by making limitation, you should considere that what will happen if x increases tow much, it means that your function's behaivior will be similar to this function: 4x/(x^2) then you can simplify this new function and you will have 4/x, which means that in very high value of X it will go to Zero, your asymptote is Y=0 and you have just one asymptote.