What does $\lim\limits_{x \to \infty} f'(x)=3$ mean?

Question

What I did was to see that the function is behaving as a straight line with gradient $3$ at infinity which implies that the function has an oblique asymptote as $x \to \infty$ but my testing portal says the answer is wrong and that instead, the function is infinite of order $1$ with respect to $x$. I think both of these are correct otherwise I must be confusing terminologies here

Answer 1

Note that in order to have a oblique asymptotes

• $y=mx+n$ with $m,n\in \mathbb{R}$.

with

$$m=\lim_{x\rightarrow+\infty}\frac{f(x)}{x}$$

and

$$n=\lim_{x\rightarrow+\infty} (f(x)-mx)$$

both limit must exist.

Note that since the function is evantually strictly increasing $f(x)\to +\infty$ thus

$$\lim_{x\rightarrow+\infty}\frac{f(x)}{x}= \lim_{x\rightarrow+\infty}f'(x)=3$$

thus $f(x)=\mathcal O(x)$.

Answer 2

The existence of an oblique asymptote is too strong. The derivative may approach 3 while the graph is not necessarily asymptotic to the line y=3x+b. It may oscillate and cross that line infinitely many times.

Answer 3

An example of a function on $\mathbb R$ with $\lim_{x\to\infty} f'(x) = 0$ but does not have a horizontal asymptote is $\log(1+x^2)$. If you prefer $f'(x) = 3$, try $$f(x) = 3x+\log(1+x^2).$$