## September 2001 #1

Problem Statement: Let $f(x)$ be differentiable on $(0,\infty)$ and suppose that $L=lim_{x\rightarrow \infty}f'(x)$ exists and is finite. Prove if $lim_{x\rightarrow \infty}f(x)$ exists and is finite, then $L=0$.

Proof: Suppose that $lim_{x\rightarrow \infty}f'(x)=L<\infty$ and that $lim_{x\rightarrow \infty}f(x)=M<\infty$. Since these limits exist we know that these limits will exist, and be the same, for any sequence $\{x_n\}$ diverging to $\infty$ so we may choose a specific sequence to look at. Let’s consider $\{x_n\}=\{2n\}$ for $n\in \mathbb{N}$. Note that $\left|x_n-x_m\right|\geq 2$ for every $n\neq m$.

Consider the sequence $\{f(x_n)\}$ which we know converges to $M$ as $n\rightarrow \infty$. Since this sequence converges, it follows that it is Cauchy.

Let $\varepsilon>0$, then there is an $N\in \mathbb{N}$ such that $\left|f(x_n)-f(x_m)\right|<\varepsilon$ for every $m>n\geq N$. Consider the closed interval $[x_n, x_{n+1}]$ for $n\geq N$. Since $f(x)$ is differentiable on $(0,\infty)$ it is also continuous on $(0,\infty)$. Since $[x_n,x_{n+1}]\subset(0,\infty)$ $f(x)$ is both differentiable and continuous on $[x_n,x_{n+1}]$ so we may use the Mean Value Theorem.

The Mean Value Theorem states that there is a $c_n\in[x_n,x_{n+1}]$ such that $f'(x_n)=\dfrac{f(x_{n})-f(x_{n+1})}{x_{n}-x_{n+1}}$. We may find such a $c_n$ in every interval $[x_n, x_{n+1}]$ and construct the sequence $\{c_n\}$ that diverges to $\infty$. Since $\{c_n\}\rightarrow \infty$ we know from our supposition that $\lim_{n\rightarrow \infty}f'(c_n)=\lim_{x\rightarrow \infty}f'(x)=L$.

So, for $n\geq N$ it follows that

$|f'(c_n)|=\left|\dfrac{f(x_n)-f(x_{n+1})}{x_n-x_{n+1}}\right|=\dfrac{|f(x_n)-f(x_{n+1}|}{|x_n-x_{n+1}|}<\dfrac{\varepsilon}{2}<\varepsilon$

But $\varepsilon$ may be made arbitrarily small, thus, $\lim_{n\rightarrow\infty}f'(c_n)=0$ which is equivalent to $\lim_{x\rightarrow\infty}f'(x)=0$. Thus, $L=0$.

$\Box$

Reflection: When trying to prove this we were getting tripped up because we were considering an arbitrary sequence $\{x_n\}$ diverging to $\infty$. The issue we had was in using the MVT, we couldn’t force $|x_n-x_m|$ to be bounded. After realizing that we already had the existence of the limit so we could choose any $\{x_n\}$ diverging to $\infty$ we were able to complete the proof. The key is that we were being asked to prove something about a limit we already knew existed, this allowed us to choose a specific sequence.