Video: Finding Limits Involving Trigonometric Functions

Find lim_(π‘₯ β†’ 0) (sinΒ²2π‘₯)/4π‘₯.

03:02

Video Transcript

Find the limit as π‘₯ approaches zero of the sin squared of two π‘₯ divided by four π‘₯.

The question is asking us to evaluate the limit as π‘₯ approaches zero of the quotient of the square of a trigonometric function and a linear function. The first thing we should do when we’re asked to evaluate a limit of this form is ask the question, are we allowed to evaluate this by direct substitution? We know we’re allowed to evaluate the limit of trigonometric functions by direct substitution. And we’re also allowed to evaluate the limit of linear functions by direct substitution. And we can also evaluate the square of a trigonometric function by direct substitution.

However, we end up with a problem when we try to evaluate the quotient of these two functions by direct substitution. When we substitute π‘₯ is equal to zero, we get the sin squared of two times zero divided by four times zero. And evaluating this expression, we get the indeterminate form zero divided by zero. So, we can’t evaluate this limit by using direct substitution. We’ll need to use a different method. We can recall a similar limit which we do know how to evaluate. We know the limit as π‘₯ approaches zero of the sin of π‘₯ divided by π‘₯ is equal to one. So, we want to rewrite the limit given to us in the question in terms of limits we do know how to evaluate. So, we’re going to try and rewrite sin squared of two π‘₯ divided by four π‘₯ in terms of sin π‘₯ divided by π‘₯.

However, we see in our function we’re taking the sin of two π‘₯, not the sin of π‘₯. And we could fix this by using the double-angle formula, and this would work. However, there’s actually a simpler method. We’ll just rewrite the limit we do know how to evaluate in terms of two π‘₯. So, we know the limit as two π‘₯ approaches zero of the sin of two π‘₯ divided by two π‘₯ is also equal to one. We do need to be careful here since the limit we’re trying to evaluate has π‘₯ approaching zero. However, our limit has two π‘₯ approaching zero. However, if two π‘₯ is approaching zero, then π‘₯ is also approaching zero. So, we can just write this limit as π‘₯ is approaching zero.

Now that we know this, let’s attempt to take a factor of the sin of two π‘₯ divided by two π‘₯ out of our function sin squared of two π‘₯ divided by four π‘₯. Our numerator is sin squared of two π‘₯, so we need another factor of the sin of two π‘₯ in our numerator. And we can see that, in our denominator, we need an extra factor of two. We can now split this limit of a product into a product of the limits. Doing this, we get the limit as π‘₯ approaches zero of the sin of two π‘₯ divided by two π‘₯ multiplied by the limit as π‘₯ approaches zero of the sin of two π‘₯ divided by two.

And we can actually evaluate both of these limits. We know the limit as π‘₯ approaches zero of the sin of two π‘₯ divided by two π‘₯ is equal to one. And we can evaluate the limit as π‘₯ approaches zero of the sin of two π‘₯ divided by two by using direct substitution. So, we substitute π‘₯ is equal to zero. This gives us the sin of two times zero divided by two. And we know the sin of zero is just equal to zero. Therefore, we’ve shown the limit as π‘₯ approaches zero of sin squared of two π‘₯ divided by four π‘₯ is equal to zero.

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