Some references show that to find an oblique asymptote of a function $f(x)$, we must see the limit of $$ m = \lim_{x \rightarrow \pm \infty} \frac{f(x)}{x} $$
If $m \ne 0$ and finite, then there is an oblique asymptote of the form $y = mx + c$.
However, I think it would be more intuitive by searching the limit of
$$ \lim_{x \rightarrow \pm \infty} f'(x) $$
If this limit exists, then we can determine the asymptote.
Question : Am I correct if I generalize the 2nd one for finding an oblique asymptote?
I have not seen any reference to use the second one (limit of $f'$) for finding an oblique asymptote. But it is more intuituive.., and we can also see from the first one that $\lim \limits_{x \rightarrow \pm \infty} \frac fx $ has an indefinite form $\frac{\infty}{\infty} $, then by L'Hopital it can be equal to $\lim f'(x)$.
Thanks in advance.
Best Answer
For a more straightforward counterexample, take $f(x) = \ln(x)$.
Its derivative $f'(x)=\frac{1}{x}$ limits to $0$ as $x \to +\infty$.
But this function has no horizontal asymptote. In fact, $\lim_{x \to \infty} \ln(x) = +\infty$ so the graph goes arbitrarily far above every horizontal line as $x \to +\infty$.
You can modify this example in many ways. For instance, $f(x) = x + \ln(x)$ has derivative limiting to $1$ as $x \to +\infty$, but the graph of $y=f(x)$ goes arbitrarily far above every slope 1 line as $x \to +\infty$, hence it has no slope 1 asymptote.