If both $\lim_{x\to\infty}f(x)$ and $\lim_{x\to\infty}f'(x)$ exist, then $\lim_{x\to\infty}f'(x)=0$.
Question. Let $f:\mathbf{R}\to\mathbf{R}$ be a differentiable function. If both the limits $\displaystyle \lim_{x\to\infty}f(x)$ and $\displaystyle \lim_{x\to\infty}f'(x)$ exist, how does one show that $\displaystyle \lim_{x\to\infty}f'(x)=0$?
Note. This statement is intuitively obvious: if $f$ has a horizontal asymptote, then the slope of the graph goes to zero as the graph approaches the asymptote.
It seems to me that proving by contradiction is easier to work with than a direct proof where one needs to estimate the size of the derivative. I will write my own answer below. Other perspectives are welcome.
1 answer
Idea. If $\displaystyle \lim_{x\to\infty}f'(x)\ne 0$, then for large enough $x$, $f'(x)$ is bounded away from $0$, which implies by the mean value theorem that $|f(x+1)-f(x)|$ is bounded away from $0$. But $|f(x+1)-f(x)|$ must also be small for large $x$ since $\displaystyle \lim_{x\to\infty}f(x)$ exists. One has a contradiction.
To formalize the idea above, here is the proof.
Proof.
Suppose $\displaystyle \lim_{x\to\infty}f(x)=L_0$ and $\displaystyle \lim_{x\to\infty}f'(x)=L$ for some real numbers $L_0$ and $L$. We show that $L=0$.
For the sake of argument, suppose $L>0$. Then it follows from the limit definition that there exists some positive real number $M>0$ such that for every $x\ge M$, $$ \frac{L}{2}<f'(x)<\frac{3L}{2}. $$ which implies (for every $x\ge M$) by the mean value theorem, $$ f(x+1)-f(x)=f'(\xi_x)>\frac{L}{2}>0,\tag{1} $$ where $\xi_x$ is some real number between $x$ and $x+1$.
On the other hand, since $\lim_{x\to\infty}f(x)=L_0$, for large enough $x$ that is greater than $M$, one has $$ |f(x+1)-f(x)|<|f(x+1)-L_0|+|f(x)-L_0|<\frac{L}{2}, $$ which contradicts (1).
Similarly, one can argue that $L<0$ is impossible.
0 comment threads
0 comment threads