Linear/Quad.

[sql_lall]

Ok, just wondering how you find the horizontal asymptote of
y = (ax + b)/(cx² + dx + e)
(i.e. a linear divided by a quadratic

I know how to do linear/linear, but not linear/quad. Is it always 0?? (if not, the book i have is bad as the only examples is horizontal asymptote = 0.)

[krtxmrtz]

The horizontal asssimptote is the limit (provided it does exist) of the derivative when x -> infinity.

The derivative of your function is (hope I haven't made any mistakes!):

y' = (-acx² - 2bcx +ae -bd) / (cx²+dx+e)²

This obviously tends to 0 for x tending to infinity regardless of the values for the parameters (a,b,c,d,e)

[sql_lall]

I take it that that means that it is always 0, which is what i thought. However, i just said (to the person asking me what the answer was) that it would always be 0, as
abs((cx² + dx + e) - (ax + b)) will always get bigger as x moves from 0, so eventually
(cx² + dx + e) will become so much bigger than (ax + b)
that
(ax + b)/(cx² + dx + e) will keep getting smaller and smaller, eventually 0 when x -> infinity

[krtxmrtz]

Well, if you want to avoid the derivatives and all the strict math then the way you put it is more elegant.
As a matter of fact, simple observation of the function shows that the denominator gets bigger at a higher rate than the numerator.

Another way to put it is: divide everything by the largest power, here x² and what's left is:

(a/x + b/x²) / (c + d/x + e/x²)

When x -> infinity, all the terms having x tend to 0 and you get 0/c = 0

The shorter and simpler the answer, the better!

Sometimes I think that using math to prove these sort of things is almost like cheating, the real thing is using one's brain!

[sql_lall]

Thanks for your answer.
I prefer it even more. (Much simpler to explain to person who asked me.)