Asymptote for $y=\frac {x^2}{kx+1}$ We want to find the asymptote of
$$y=\frac {x^2}{kx+1}$$
as $x\to \infty$.
By synthetic division or binomial expansion, we arrive at $$y=\frac 1k \left(x-\frac 1k\right)$$
which is the correct answer.
However we could also have divided top and bottom by $x$ to get:
$$y=\frac x{k+\frac 1x}$$
and conclude that as $x\to\infty$, $\frac 1x \to 0$, hence $y\to \dfrac kx$ which is close but not the right answer. How do we reconcile this method with the ones mentioned earlier?
Thanks.
 A: Your second analysis is not accurate. You cannot just remove $1/x$ because it goes to zero. You need to find the second term in the expansion at infinity, in this case the constant term. You can do
$$
y=x(k+1/x)^{-1}=\frac x{k}(1+1/kx)^{-1}=\frac x{k}(1-1/kx+\dots)=\frac 1{k}(x-1/k)+\dots
$$
A: You're misinterpreting what it means to find the asymptote of a function. It's NOT the same as finding the limit of the function.
And in your second attempt, you apparently set out to find the limit of the function, $\lim\limits_{x\to\infty}y(x)$ — which, as I said, is not the same thing, and which you didn't do correctly. Let me reiterate: once you said that you want to see what happens to $y$ as $x\to\infty$, you're talking about $\lim\limits_{x\to\infty}y(x)$. But if you're finding this limit with respect to $x$, you can't leave some of the $x$'s still in the "answer". As $x\to\infty$, not only does $1/x\to0$ in the denominator, but also $x\to\infty$ in the numerator; and as a result we find that
$$\lim_{x\to\infty}y(x)=\lim_{x\to\infty}\frac{x}{k+\frac{1}{x}}=\frac{\infty}{k+0}=\infty,$$
which is certainly true, but not particularly useful.
Finding an asymptote for a function $y(x)$ means that we need to find a linear function $ax+b$ such that $\lim\limits_{x\to\infty}[y(x)-(ax+b)]=0$. Your second approach doesn't address this question.
A: Aim:
Find the best linear approximation $y=mx+b$ to the function $y=f(x)$ for large $x: $
1)The second option gives
$y=mx$, where $m=(1/k)$, which is a linear approximarion for large $x$
2)A better linear approximation of
$y=(1/k)x(1+1/(kx))^{-1}=$
$(1/k)x(1-1/(kx)+O(1/x^2))$
is given by
$y=(1/k)(x-1/k)$, where you retain the constant term.
A: Your second analysis merely hints that the curve behaves like a straight line at infinity -- it doesn't give you its exact form. This is as a result of the fact that the analysis is quite crude.
For exact analysis, write the expression as $$\frac{x^2}{kx}\left(\frac{1}{1+1/kx}\right)=\frac xk\left(1-\frac{1}{kx}+\frac{1}{k^2x^2}-\cdots\right),$$ from where the exact form of the linear asymptote becomes clear.
