Video: The Derivative as a Limit of a Function

Find lim_(β„Ž β†’ 0) (cot ((πœ‹/3) + β„Ž) βˆ’ cot (πœ‹/3))/β„Ž.

02:49

Video Transcript

Find the limit as β„Ž approaches zero of cot of πœ‹ by three plus β„Ž minus cot of πœ‹ by three all over β„Ž.

Now actually, believe it or not, what we’re not going to do is evaluate this limit using our usual methods. Instead, we’re going to recall the definition of the derivative. The derivative of a function 𝑓 of π‘₯ with respect to π‘₯ at a point π‘₯ equals π‘Ž is given by 𝑓 prime of π‘Ž. And it’s defined as the limit as β„Ž approaches zero of 𝑓 of π‘Ž plus β„Ž minus 𝑓 of π‘Ž over β„Ž. We’re going to compare the limit in this question with our definition for the derivative. When we do, we see that 𝑓 of π‘Ž in our question is cot of πœ‹ by three. And this means 𝑓 of π‘₯ is equal to the cot of π‘₯.

Now that we’ve defined 𝑓 of π‘₯, it’s clear we need to find the derivative of our function 𝑓 prime of π‘₯ and then evaluate that at π‘Ž, at πœ‹ by three. So, what is the derivative of cot of π‘₯? Well, it’s helpful to recall that cot is equal to one over tan. And since tan is sin of π‘₯ over cos of π‘₯, one over tan of π‘₯ is cos of π‘₯ over sin of π‘₯. We should now be able to see that 𝑓 of π‘₯ is the quotient of two differentiable functions, so we can use the quotient rule to find its derivative. This tells us that for two differentiable functions 𝑒 and 𝑣, the derivative of 𝑒 divided by 𝑣 is 𝑣 times d𝑒 by dπ‘₯ minus 𝑒 times d𝑣 by dπ‘₯ all over 𝑣 squared.

Well, we can see that 𝑒 here is equal to cos of π‘₯ and 𝑣 is equal to sin of π‘₯. Differentiating 𝑒 with respect to π‘₯, that’s the derivative of cos of π‘₯, we get negative sin of π‘₯. Then d𝑣 by dπ‘₯, that’s the derivative of sin of π‘₯, is cos of π‘₯. So then, 𝑓 prime of π‘₯, the derivative, is 𝑣 times d𝑒 by dπ‘₯, that’s sin of π‘₯ times negative sin of π‘₯ which is negative sin squared π‘₯, minus 𝑒 times d𝑣 by dπ‘₯, that’s cos of π‘₯ times cos of π‘₯ which is cos squared π‘₯, and that’s over 𝑣 squared, so that’s over sin π‘₯ all squared or sin squared π‘₯.

Next, we need to recall that sin squared π‘₯ plus cos squared π‘₯ is equal to one. And then, if we factor negative one on our numerator, we get negative one sin squared π‘₯ plus cos squared π‘₯, which means that 𝑓 prime of π‘₯ is negative one over sin squared π‘₯. Now, we could, of course, write that as a negative csc squared π‘₯. And we could quote that the general result for the derivative of cot of π‘₯ is negative csc squared π‘₯. But the process is quite straightforward, so it’s good to see where it comes from. And so this means, to find the limit as β„Ž approaches zero of cot of πœ‹ by three plus β„Ž minus cot of πœ‹ by three over β„Ž, we’re going to evaluate the derivative of our function at πœ‹ by three.

That’s negative one over sin squared of πœ‹ by three. The sin of πœ‹ by three is root three over two. So, 𝑓 prime of πœ‹ by three is negative one over root three over two all squared. That’s negative one over three over four. And of course, when we divide by a fraction, it’s the same as multiplying by the reciprocal of that fraction. So, this becomes negative one times four over three, which is negative four over three. And since we said that our original limit was equal to the derivative evaluated at πœ‹ by three, we know that our original limit is equal to negative four-thirds.

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