Is the method of finding a slant asymptote correct?

Question

I understand the idea behind finding a slant asymptote: we asume, that at an infinite distance of argument, function should be equal to some line function, so to check that we take a line function expression that is $kx+b$, set it equal to inspecting function expression, and find a limit of it, where argument goes to infinity.

In sources I read, it is done in illegal way: inspecting function is set to be equal to just $kx$, from where $k$ is obtaining as $k=\lim_{x \to \infty} \dfrac{f(x)}{x}$, and after $k$ is obtained, it just passed to complete line expression $kx+b=f(x)$ as a constant, and now $b$ is obtaining in the same way: $b=\lim_{x \to \infty}f(x)-kx$.

Actually, I can't surely state why it is wrong, but, probably, because $k$ in $kx + b$ is just an another variable?

For example, we obviously can't solve $3x+y=1$, by just removing $y$ for a while, as $3x=1,\space x=\dfrac{1}{3}$ and then put it back, getting $\dfrac{3}{3}+y=1, \space y = 0$... And when I ended typing it understand, that $x=1/3$ and $y=0$ are solution for the equation above...

I am totally confused, can someone explain where my algebra knowledges are wrong?

Answer 1

The reason you can separate out the two parts is because of what the limit does to the constant $b$. Suppose that $\lim{x \rightarrow \infty} = kx+b$. Then,

$$\lim_{x \rightarrow \infty}\frac{kx+b}{x} = k+\lim_{x \rightarrow \infty}\frac{b}{x}.$$

The second limit is zero, so that only $k$ is left. Now, notice that $$\lim_{x \rightarrow \infty} (f(x)-kx) = \lim_{x \rightarrow \infty} ((kx+b)-kx) = b.$$

Answer 2

First of all, it is better to be precise with the definitions. We say that $f(x)$ has a slant asympote for $x \to +\infty$ if, for some $m, q \in \mathbb R$, with $m\neq 0$, we have $$\lim_{x \to +\infty} [f(x)-mx -q] = 0.\tag{1}\label{1}$$ It is not always easy to understand if $m$ and $q$ exist and to find them, by directly using \eqref{1}.

The idea behind the procedure you mention is to notice that if \eqref{1} is true, then $$\lim_{x\to + \infty} \frac{f(x)}x = m.\tag{2}\label{2}$$ So \eqref{2} is a necessary condition for \eqref{1} to hold true. Therefore you can first verify if \eqref{2} holds true for some $m\neq 0$. And then use this value of $m$ in \eqref{1} and verify whether the limit $$\lim_{x \to +\infty}[f(x)-mx]$$ exists and it is finite.

---

EDIT. Proof that \eqref{2} follows from \eqref{1}.

If \eqref{1} is true, then $$\lim_{x\to +\infty} [f(x) - mx] = q.$$ Thus $$\lim_{x\to +\infty}\frac{f(x)-mx}{x} = 0,$$ which can be written as $$\lim_{x\to +\infty}\left[\frac{f(x)}x - m\right] =0,$$ which in turn is equivalent to \eqref{2}.

---

EXAMPLE 1.

In some cases it is easy to find $m$ and $q$ directly by inspecting the function. For example $$f(x) = \frac{x^2-2x}{x+1}$$ can be written as \begin{eqnarray} f(x) &=&\frac{x^2+x-3x}{x+1}\\ &=& \frac{x(x+1)-3x-3+3}{x+1} =\\ &=& \frac{x(x+1)-3(x+1)+3}{x+1} = \\ &=& x-3 + \frac{3}{x+1}, \end{eqnarray} which shows that \eqref{1} is true for $m=1$ and $q=-3$.

EXAMPLE 2.

There are cases, though, in which using \eqref{2} is very convenient, in order to find the equation of the slant asymptote, because it splits the problem in two parts. Take $$f(x) = (x-3) e^{\frac{x-1}{x-2}}.$$ First observe that when $x\to \infty$ the exponent converges. This hints at the possiblity that \eqref{2} might be true for some $m\neq 0$. In fact we have $$\lim_{x\to \infty} \frac{f(x)}x = e$$ (exercise for you). We still do not know if $f$ has a slant asyptote, though, since \eqref{2} is a necessary (not sufficient!) condition. But now we can see if there is a $q$ for which \eqref{1} holds true. I.e. we check the limit $$\lim_{x\to \infty}[f(x)-ex].$$ I leave it as an exercise for you to find that the limit in fact exists and is equal to $-2e$, so that we have a slant asympote with equation $$y=e(x-2).$$