Can we determine an oblique asymptote of a function by the limit of $f'(x)$? 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.
 A: Let $f(x)=x+\sin{\sqrt{x}}.$ Then $f'(x)=1+\dfrac{\cos{\sqrt{x}}}{2\sqrt{x}} \underset{x\to+\infty}{\to} 1,$ but oblique asymptote does not exists since $\nexists \lim\limits_{x\to+\infty}(f(x)-x).$ 
Another example: 
The function $g(x) = x+\dfrac{\sin{x^2}}{x}$ has an asymptote $y=x,$ but
it's derivative $g'(x)=1+2\cos{x^2}-\dfrac{\sin{x^2}}{x^2}$ does not have a limit as $x\to\infty.$
A: 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.
A: This works for rational functions, but not for more general functions. For example,
$$ \lim_{x \to \infty} \frac{x+\sin{x}}{x} = 1, $$
but the derivative of $x+\sin{x}$ is $1+\cos{x}$, which does not have a limit as $x \to \infty$.
