**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.

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