Video: Finding Limits Involving Trigonometric Functions

Alex Cutbill

Evaluate lim_(π‘₯ β†’ 0) (sin π‘₯)/(sin (π‘₯/2)).

05:18

Video Transcript

Evaluate the limit of sin π‘₯ over sin π‘₯ over two as π‘₯ approaches zero.

Let’s take a look at this limit. The first thing we should try is direct substitution. We substitute the number that π‘₯ is approaching, zero, for π‘₯ in the expression. And doing this, we get sin zero over sin zero over two. The sin of zero is zero, so have a zero in the numerator. And also of course sin of zero over two is just sin of zero, which is zero. And so we have zero in the denominator as well.

Zero over zero is an indeterminate form. And so unfortunately direct substitution hasn’t given us the value of the limit that we’re looking for. Still it was worth a try. So how do you evaluate this limit? Well we’re going to see two methods in this video. The first method makes use of the double-angle identity for sine, that sin two πœƒ is two times sin πœƒ times cos πœƒ for any angle πœƒ.

And if we choose to substitute π‘₯ over two for πœƒ, we find that sin of two times π‘₯ over two is two sin π‘₯ over two times cos π‘₯ over two. And of course on the left-hand side, two times π‘₯ over two is just π‘₯. So we find that sin π‘₯ is two sin π‘₯ over two cos π‘₯ over two. Okay, but how does this help us? Well, if we use the expression in our limit, we see that we have a factor of sin π‘₯ over two in the numerator, which will cancel with the sin π‘₯ over two in the denominator. Upon the cancelling, what do we get?

We get the limit of two cos π‘₯ over two as π‘₯ approaches zero. And this is a limit that we can evaluate using direct substitution. Replacing π‘₯ by zero, we get two cos zero over two. And cos zero over two is just cos zero, which is one. And so our answer is two times one, which is two. This method required using a clever substitution for sin π‘₯: a substitution which may not have been obvious, but once we’ve made it, it was clear what to do.

We can however use a different trick to evaluate this limit. This method uses the fact that the limit of sin π‘₯ over π‘₯ as π‘₯ approaches zero is one. This is an important limit as it allows us to differentiate sine. If you haven’t seen this limit before, then it’s probably best to stop the video here and make do with the first method. But if you have seen this limit before, you might be interested in how it can help us solve our problem. Well we can write sin π‘₯ over sin π‘₯ over two as sin π‘₯ over π‘₯ times π‘₯ over sin π‘₯ over two.

You can check that this is right by multiplying out the fractions on the right-hand side and then simplifying. Basically what we’ve done is to divide and then multiply by π‘₯. And we can then use the fact that the limit of a product of functions is the product of their limits. And so we can write this limit as the product of two limits. And one of these limits we know the value of: the limit of sin π‘₯ over π‘₯ as π‘₯ approaches zero is one. And so we only have the other limit to worry about: the limit of π‘₯ over sin π‘₯ over two as π‘₯ approaches zero.

How do we find this limit? Well it’s very nearly the reciprocal of the limit we know. The limits of the reciprocal of some function is just the reciprocal of the limit of that function. And so we see that the limit of π‘₯ over sin π‘₯ as π‘₯ approaches zero is one. And there’s nothing particularly special about the use of the letter π‘₯ here. We could use any letter, perhaps πœƒ. Or thinking of the limits we have, we could set πœƒ equal to π‘₯ over two. The limit of π‘₯ over two over sin π‘₯ over two as π‘₯ over two approaches zero is one.

You might feel slightly uneasy about replacing πœƒ by something that isn’t just another letter, but this can be justified. What does it mean for π‘₯ over two to approach zero? Well this just means that π‘₯ itself is approaching zero. And so we have that the limit of π‘₯ over two over sin π‘₯ over two as π‘₯ approaches zero is one. This is really close to the limit we have. The only difference being that we have π‘₯ in the numerator rather than π‘₯ over two. But we can write that π‘₯ in the numerator as two times π‘₯ over two, and we know that the limit of two times something is two times the limit of that something.

Now we have a limit whose value we know. The value of this limit is one. The value of the limit in our question is therefore two times one, which is two. And this is our final answer. This method didn’t require using a clever identity for sin π‘₯. In general, using the limit of sin π‘₯ over π‘₯ as π‘₯ approaches zero and the corresponding limits for cosine and tan allows you to rewrite a trigonometric limit as a limit of a rational function, which you may find easier to solve.

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