Problem Statement: Let be differentiable on and suppose that exists and is finite. Prove if exists and is finite, then .
Proof: Suppose that and that . Since these limits exist we know that these limits will exist, and be the same, for any sequence diverging to so we may choose a specific sequence to look at. Let’s consider for . Note that for every .
Consider the sequence which we know converges to as . Since this sequence converges, it follows that it is Cauchy.
Let , then there is an such that for every . Consider the closed interval for . Since is differentiable on it is also continuous on . Since is both differentiable and continuous on so we may use the Mean Value Theorem.
The Mean Value Theorem states that there is a such that . We may find such a in every interval and construct the sequence that diverges to . Since we know from our supposition that .
So, for it follows that
But may be made arbitrarily small, thus, which is equivalent to . Thus, .
Reflection: When trying to prove this we were getting tripped up because we were considering an arbitrary sequence diverging to . The issue we had was in using the MVT, we couldn’t force to be bounded. After realizing that we already had the existence of the limit so we could choose any diverging to 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.