Suppose $f,g: \mathbb R \rightarrow \mathbb R$, have linear growth or slower (i.e. $\exists \delta,M_1,M_2>0$ such that for $x>\delta$, $|f(x)| \leq M_1 x$ and $|g(x)| \leq M_2 x$) and $g$ grows at linear rate less than 1 (i.e. $M_2<1$).

Further suppose $\lim_{x\rightarrow \infty} f(x)-f'(x)g(x)=d$ for some finite $d$ and $\lim_{x \rightarrow \infty} f'(x)=c$ for some finite $c$.

Is it possible to show that $\lim_{x\rightarrow \infty} f(x)-cg(x)=d$. I am most interested in showing this in the case when $\lim_{x\rightarrow \infty} g(x)=\infty$

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    $\begingroup$ Could you please add some context? Like , where are you looking to apply this result? Or is it available in some textbook? I'd love to know what motivated you (especially, why the sub-linear growth) and finally : $g$ isn't differentiable, right? $\endgroup$ Nov 19, 2021 at 18:02
  • $\begingroup$ If $g$ differentiable helps to show the result, then that is ok to assume. Or even the case of $g(x)$ is linear in $x$ (i.e. $g(x)=\alpha x$) would be helpful to understand. $\endgroup$
    – Jong
    Nov 19, 2021 at 20:18
  • $\begingroup$ I think the key to the problem is showing $(c-f'(x))g(x) \to 0$, i.e. $f'(x)$ approaches $c$ faster than $x^{-1}$. I have not succeeded in proving that yet. $\endgroup$
    – Xiaoyu Liu
    Nov 26, 2021 at 6:12

1 Answer 1


Start with the knowledge that calculating with known limits is allowed.

Our known limits are




$c$ can than be rewritten as


And must just be rewritten to state the equation under proof. And that is allowed plainly by knowing we can calculate with limits.

Reworked this with the given bounding example the linear function. In this case, taking the limit of the derivative $f'(x)$ is not very meaningful because $f'(x)$ is a constant on ℝ. With the upper boundary, the rest gets as simple as calculating. $g(x)$ is always a shifted $f(x)$. Both are separated by a constant.

What about the case were $\lim_{x\rightarrow\infty}g(x)=\infty$. Now the boundary example of a linear function is too $\infty$ for large values but more can not happen to obey the initial constraint.

Get this more certain. $f(x)-g(x)$ is smaller growing or decreasing than $(M_{1}-M_{2})x$ even absolute. This differential of the difference function is a constant.

This first given limit poses a much tougher constraint on $g(x)$ together with the second limit for $f'(x)$. You can easily attempt an assumption that for some $\delta$ $f'(x)$ is not approaching $c$ than the first limit $d$ would be different is the consequence. This is seen by rewritting the limit function under consideration again to $f'(x)$.

From the representation of c as a quotient inversely proportional to $g(x)$ follows that a larger growth of $g$ induces smaller $c$. In general cases $g(x)$ can only differ very little from $f(x)$ so that $g(x)\approx \frac{f(x)-d}{f'(x)}$. That is like a construction receipe for $g(x)$.

There is a nice calculator just for limits on the web: limit-calculator.

The absolute slower growth of the functions open the functions the negative range. Calculating with absolute values should be knowledge, working knowledge as well at the time such problems are given to students. This is simplified by the fact that both functions vary absolute slower than linear.


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