If $\displaystyle \lim_{x \to \infty} f(x)=L$ and $\displaystyle \lim_{x \to \infty} f'(x)$ exists, then $\displaystyle \lim_{x \to \infty}f'(x)=0$

Using contradiction, I proved the following statement:

If $\displaystyle \lim_{x \to \infty} f(x)=L$ and $\displaystyle \lim_{x \to \infty} f'(x)$ exists, then $\displaystyle \lim_{x \to \infty}f'(x)=0$

I'd like to know if there is a more standard approach as mine seems kind of messy.

Suppose $\displaystyle \lim_{x \to \infty} f'(x)=K \neq 0$. WLG, assume $K \gt 0$. By assumption, we know that there is a $N_{f',\frac{K}{2}}$ such that for any $x \in \left(N_{f',\frac{K}{2}}, \infty\right)$, we have $|f'(x)-K| \lt \frac{K}{2}$, which means $f'(x) \gt \frac{K}{2} \gt 0$ $\quad (\dagger)$.

As such, for any $x_2,x_1 \in \left(N_{f',\frac{K}{2}}, \infty\right)$, we can apply the Mean Value Theorem on the continuous interval $[x_1,x_2]$...the interval is continuous because $f'(x_1)$ and $f'(x_2)$ are necessarily defined. The MVT and $(\dagger)$ show that for some $x^* \in (x_1,x_2)$: $\frac{f(x_2)-f(x_1)}{x_2-x_1} = f'(x^*) \gt \frac{K}{2}$, which implies that $f(x_2) \gt \frac{K}{2} (x_2-x_1)+f(x_1) $. In particular, this shows that $f(x_2) \gt f(x_1)$.

Next, by $(\dagger)$ we know that if we go sufficiently far away to the right, $f'$ has a positive floor value of $\frac{K}{2}$. This means if we go out far enough, we can find a $c$ such that $f(c) \gt L$. To see this, pick any arbitrary $x \in \left(N_{f',\frac{K}{2}}, \infty\right)$. This is a reference value. Next, solve the following equation for $c$: $\frac{K}{2}(c-x)+f(x)=L$

Finally, consider $\varepsilon = f(c)-L$. By assumption, there is an $N_{f,\varepsilon}$ such that for any $x \in \left(N_{f,\varepsilon},\infty \right)$, we have that $|f(x)-L| \lt f(c) - L\quad (\dagger \dagger)$.

Let $N = \max(N_{f,\varepsilon},c)$. Consider any $z \in (N, \infty)$. Importantly $z \gt c$. By $(\dagger)$, we have that $f(z) \gt \frac{K}{2}(x-c)+f(c)$, which implies that $f(z) \gt f(c)$.

However, by $(\dagger \dagger)$ we know that $|f(z)-L| \lt f(c)-L$, but this implies that $f(z) \lt f(c)$, a contradiction. $\quad \square$

Solution 1:

You could apply the mean value theorem at an interval of the form $[x, x+1]$. This guarantees the existence of a $\xi_x \in (x, x+1)$ such that $f'(\xi_x) = f(x+1)-f(x)$. Since the limit of $f'(x)$ exists at infinity we get that $\lim_{x\to \infty} f'(x)=0$ by taking limits on both sides and noting that $\xi_x \to \infty$ as $x\to \infty$.