Is there a grid that came with the problem?
1/2 &50 0.5 hope this helps
The answer is pretty simple, 1/8, 1/5, 1/4, 1/2
We have

So

The rational term vanishes as <em>x</em> gets arbitrarily large, so we can ignore that term, leaving us with

and this happens if <em>a</em> = 1 and <em>b</em> = -1.
To confirm, we have

as required.