Proving an implication of the form: $\lim_{x \to \infty}$ of derivative/function implies $\lim_{x \to \infty}$ function/derivative?

Question

(Assume $?$ is some function that is continous and differentiable everywhere)

For example, if $\lim_{x \to \infty} ?' = 7$ implies $\lim_{x \to \infty} ? = \infty$

or $\lim_{x \to \infty} ? = 5$ implies $\lim_{x \to \infty} ?' = something$

Basically, how do you (or how would you) start writing a proof that is something along the lines of:

$\lim_{x \to \infty} ?' = something$ implies $\lim_{x \to \infty} ? = something$

$\lim_{x \to \infty} ? = something$ implies $\lim_{x \to \infty} ?' = something$

TL;DR How would you start to think about this type of problem (i.e. limit of derivative/function implies limit function/derivative) and formulate a plan and prove the question? I cannot think of a good starting point to snowball into some kind of solution since the definition of limit doesn't get me anywhere.

I have thought about using L'Hopital's rule shown here, but you would have to show $\lim_{x \to \infty} \frac{f(x)}{x} = a$ implies $f(x)→\infty$ for $x→\infty$ to be rigourous, but that seems like an equally difficult problem. If $a = 0$ it does not hold (as it says a, a must be greater than zero), but under what circumstances can you assume $\lim_{x \to \infty} \frac{f(x)}{x} = a$ and that $a$ must be greater than zero?

Some other proofs of similar questions use MVT but also Cauchy's subsequence and a lot of different messy variables and inequalities, which is much more complicated then these problems should require (for a 'simple' analysis question) I think (or maybe not, I'm not sure). The last proof here only uses MVT but the conclusion doesn't make much sense to me ($f(y)>M$), and the post is so old that commenting is probably pointless.

To me it seems like using the definition of limit is the best place to start, such as trying to show $\forall \varepsilon > 0$, $\exists M>0$ such that $x>M$ implies $\mid f(x) - L \mid < \varepsilon$ for some limit as $x$ approaches infinity and equals some number $L$. If $L = \infty$ then perhaps showing $\forall M > 0$, $\exists N>0$ such that $f(x)>M, \forall x>N$ would work as well. But that is the hard part that I get stuck on.

My problem is that I do not see how MVT or L'Hopital's Theorem (The most useful theorems I could think of to solve the problem) or the given assumption that $\lim_{x \to \infty} ? = something$ or $\lim_{x \to \infty} ?' = something$ (depending on the problem, since usually the if part is used to prove the then part) could be used to show that $f(x)>M$, $\forall x>N$ or $\mid f(x) - L \mid < \varepsilon$ etc depending on what $something$ is equal to and what $?$ is. To me it seems impossible to prove with the limited amount of information given (which it is not, but it seems like it), which leaves me stuck.

Answer 1

In the first case, the definition is indeed a good way to start. If $a > 0$ and $\lim_{x \to \infty}f'(x) = a$, then there exists $N \in \mathbf R$ with the property that $x > N$ implies $f'(x) > \frac a2$. Now it is a matter of using the mean value theorem. If $x > N$ then there is a point $c$ in between $x$ and $N$ (and in particular larger than $N$) satisfying $$f(x) - f(N) = f'(c)(x-N) \ge \frac a2 (x-N)$$ so that $$x > N \implies f(x) > \frac {ax}2 - \frac{aN}{2} + f(N).$$ This leads quickly to $f(x) \to \infty$. A corresponding conclusion can be reached if $a < 0$.

The other direction doesn't work so well. If $\lim_{x \to \infty} f(x) = a$ you can conclude that if $f'$ has a limit it must be $0$, but there is no particular reason for that limit to exist.

Answer 2

Your first conjecture that if $\lim_{x\to\infty} f'(x)=7$ then $\lim_{x\to\infty} f(x)=\infty$ is correct (and we can also replace $7$ by any positive number). The way to think of this is to imagine a function that satisfies these conditions. As $x$ gets bigger and bigger the function behaves (i.e. grows) like a straight line with slope $7$. It thus makes sense to conjecture that indeed the function will tend to infinity.

To prove this we can note that since $\lim_{x\to\infty} f'(x)=7$ then in particular by the definition of a limit there is some $x_0$ such that for all $x>x_0$ we have $f'(x)>6$. Now if we take any $x>x_0$ then by the MVT there is some $c\in (x_0,x)$ such that $\frac{f(x)-f(x_0)}{x-x_0}=f'(c)>6$, that is, $f(x)>6(x-x_0)+f(x_0)$. The right hand side tends to infinity therefore the left hand side does as well.

Your second question is if by knowing $\lim_{x\to\infty} f(x)=5$ we can know anything about $\lim_{x\to\infty} f'(x)$. Naively one might be tempted to conjecture that since the function tends to $5$ then its derivative might tend to $0$, but this is incorrect; the function can oscillate around the value $5$ while still tending to it. For instance, the function $f(x)=5+\frac{\sin(x^2)}{x}$ tends to $5$ but it is a simple calculation to see that its derivative does not tend to $0$ (it does not converge to any value as $x\to\infty$).

This doesn't mean we can't say anything at all about $f'(x)$ in this case; we can say that we know for sure that $\lim_{x\to\infty} f'(x)$ is either $0$ or else it doesn't exist, because if the limit was a positive number then by what we showed above we'd have that $\lim_{x\to\infty} f(x)=\infty$.