Video Transcript
Find the limit of Δ𝑥 tends to zero of cos of 𝜋 by six plus Δ𝑥 minus cos of 𝜋 by six all over Δ𝑥.
Here, Δ𝑥 is just a variable. So if we liked, we could rewrite our limit using another variable, say ℎ in place of Δ𝑥. We’ll just stick with Δ𝑥 in this video. If we try to evaluate this limit using direct substitution of Δ𝑥 equals zero, then we obtain the indeterminate form zero over zero, which is undefined. So we cannot use direct substitution to evaluate this limit. There are however a variety of other methods that we can use. In this video, we will look at a method which involves using the definition of derivative. Recall that the definition of the derivative of a function 𝑓 is the limit of Δ𝑥 tends to zero of 𝑓 of 𝑥 plus Δ𝑥 minus 𝑓 of 𝑥 all over Δ𝑥. This looks a bit similar to the limit we are asked to evaluate in the question.
Letting 𝑓 of 𝑥 equal cos of 𝑥, we can rewrite the limit in the question as follows. If cos of 𝑥 equals 𝑓 of 𝑥, then cos of 𝜋 by six plus Δ𝑥 equals 𝑓 of 𝜋 by six plus Δ𝑥. And cos of 𝜋 by six equals 𝑓 of 𝜋 by six. Now, our limit looks more similar to the definition of derivative. In fact, our limit is equal to the derivative of 𝑓 evaluated at 𝜋 by six. Since 𝑓 of 𝑥 equals cos 𝑥, the derivative of 𝑓 is negative sin 𝑥. So the derivative of 𝑓 evaluated at 𝜋 by six is negative sin of 𝜋 by six.
Let’s work out the numerical value of negative sin of 𝜋 by six. We can work this out by drawing the equilateral triangle with all sides of length two, cutting it in half. And you think the ratio sin of 𝜃 equals opposite over hypotenuse with 𝜃 equals 𝜋 by six. Doing so, we obtain that sin of 𝜋 by six equals one divided by two, i.e., a half. So negative sin of 𝜋 by six equals negative one-half, which is the final answer.
