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.

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This entry was posted in Analysis, Cauchy, Continuity, Differentiable, Math, MVT, Sequence and tagged , , . Bookmark the permalink.

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