# Question Video: Finding the One-Sided Limit of Trigonometric Functions at a Point Mathematics

Determine the following infinite limit: lim_(π₯ β 3πβΊ) β5 cot 2π₯.

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### Video Transcript

Determine the following infinite limit: The limit as π₯ approaches three π from the right of negative five times the cot of two π₯.

In this question, weβre asked to evaluate the limit of a trigonometric function. And weβre told that this limit evaluates to give us an infinite value. Since our values of π₯ are approaching three π from the right, this means the outputs of the function are approaching either positive or negative β as our values of π₯ get closer and closer to three π from the right. So π₯ is greater than three π.

We need to determine which of these two cases weβre in. Do the outputs increase without bounds or do they decrease without bound? And thereβs a few different things we could try. For example, we could construct a table of values by substituting values of π₯ into our function. And then we could analyze the output values as π₯ approaches three π from the right. This would give us an idea of what happens to the outputs of the function. However, a table of values will only give us an idea of whatβs happening. It wonβt be enough to prove what the limit evaluates to.

So weβll need to try and evaluate this limit by using a different method. Letβs start by rewriting our limit by using the following trigonometric identity. We know that the cotangent is one divided by the tangent, so we can rewrite cot of two π₯ as one divided by tan of two π₯. Therefore, we can rewrite our limit as the limit as π₯ approaches three π from the right of negative five divided by the tan of two π₯. And this then gives us a few different ways of evaluating this limit. For example, we could sketch the graph of negative five divided by the tan of two π₯, which is, of course, the same as the graph of negative five times the cot of two π₯.

And one way to do this is to recall the cot of π₯ has vertical asymptotes at integer multiples of π. We can use this to sketch cot of π₯, where we want to take particular attention to what happens at each of the vertical asymptotes. If π₯ approaches one of the vertical asymptotes from the right, we approach β. And if π₯ approaches one of the vertical asymptotes from the left, we approach negative β. Keeping this in mind, we can use this to sketch the graph of cot of two π₯. Weβre multiplying the input values by two, which means weβre stretching our graph horizontally by a factor of one-half.

Finally, we need to multiply this value by negative five, which means we stretch it vertically by a factor of five and reflect it in the horizontal axis. We can apply both of these transformations at once to sketch the graph of π¦ is equal to negative five cot of two π₯. Well, thereβs a few things worth pointing out here. First, we now have vertical asymptotes at every integer multiple of π by two. This is because we stretched our graph horizontally by a factor of one-half. Next, the behavior of the asymptote has flipped. This is because weβve reflected our graph in the horizontal axis. Now, when we approach one of our asymptotes from the right, we approach negative β. And when we approach one of the asymptotes from the left, we approach positive β.

Finally, because the tangent function and the cotangent function are periodic, we know we donβt have to graph the entire function because the graph will repeat itself. The tangent function has a period of π, so the tangent of two π₯ will have a period of π by two. This then gives us two options. We can use this to continue sketching the rest of our graph. Or we can just evaluate the limit as π₯ approaches three π from the right of this function by using the fact itβs periodic. It will be the same as the limit as π₯ approaches zero from the right. And this is of course just saying the behavior of the graph around the asymptotes is the same at all of the asymptotes.

And we can then just evaluate this limit from the graph. As π₯ approaches zero from the right, our output values are decreasing without bound. They approach negative β. Therefore, the limit as π₯ approaches three π from the right of negative the cot to two π₯ is negative β. And we could leave our answer here. However, we wonβt always be able to graph the function weβre evaluating the limit of. So letβs go through a method where we donβt need to sketch the graph.

Instead, letβs go back to our limit. We can see the numerator of our limit remains constant. Itβs always equal to negative five. However, our denominator is the tan of two π₯. Of course, we know as π₯ approaches three π from the right, the tan of two π₯ will approach the tan of six π, which is zero. This is just another way of saying we can evaluate trigonometric functions by using direct substitution. However, we can actually determine the value of this limit by considering what the outputs of the tan of two π₯ are around π₯ is equal to three π.

And although itβs not necessary, weβre going to do this in a very similar way we did for the cotangent function. We sketch π¦ is equal to the tan of two π₯, which is just the tan of two π₯ stretched horizontally by a factor of one-half. And since tan of π₯ has its vertical asymptotes at π by two plus integer multiples of π, the graph of tan of two π₯ will have its vertical asymptotes at π by four plus integer multiples of π by two.

However, we donβt actually need this information. We want to see what happens to this graph as π₯ approaches three π from the right. And to do this, we just note the π₯-intercepts of this graph will be the integer multiples of π by two. Because if we substitute an integer multiple of π by two into this function, we get two times the multiple of π by two, which is a multiple of π. So this is the tangent of a multiple of π. Now we can use exactly the same argument we did before. This graph is periodic with a period of π by two. So π₯ approaching three π from the right will be the same as π₯ approaching zero from the right.

Now we might conclude here that this is approaching zero. However, we want to note the sign. The sign is always positive. And this gives us a very useful piece of information about this limit. The numerator remains constant. However, the denominator is a smaller and smaller positive number. And a negative number divided by a positive number is always a negative number. Therefore, this is decreasing without bounds. Its limit is negative β. Therefore, we were able to show the limit as π₯ approaches three π from the right of negative five times the cot of two π₯ is negative β.