Video: Finding Limits Involving Trigonometric Functions

Find lim_(π‘₯ β†’ 0) 4π‘₯Β²/sinΒ² 5π‘₯.

04:15

Video Transcript

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

We see the question is asking us to evaluate the limit as π‘₯ approaches zero of the quotient of two functions. In this case, it’s the quotient of a polynomial function and the square of a trigonometric function. The first thing we should think about when asked to evaluate a limit such as this is, are we allowed to use direct substitution?

And in this case, we can attempt to evaluate this by using direct substitution since we can evaluate polynomials and the squares of trigonometric functions by direct substitution. Substituting π‘₯ is equal to zero, we get four times zero squared divided by the sin squared of five times zero. And if we evaluate this expression, we get the indeterminate form zero divided by zero. And this does not mean we can’t evaluate this limit.

What this tells us is, we cannot determine the value of this limit by using this method. So we’re going to need to think of a different way to try and evaluate this limit. For example, we could try rewriting this limit in terms of limits we do know how to evaluate. For example, we know the standard trigonometric limit result that the limit as π‘₯ approaches zero of the sin of π‘₯ divided by π‘₯ is equal to one. And this is similar to the limit given to us in the question.

However, in that case, we have π‘₯ in our numerator and our sine function in the denominator. Luckily, we know a result about limits which will let us take the reciprocal of the function inside of our limit. We know if the limit as π‘₯ approaches π‘Ž of some function 𝑓 of π‘₯ is equal to 𝐿, then the limit as π‘₯ approaches π‘Ž of the reciprocal of 𝑓 of π‘₯ is equal to one divided by 𝐿. And of course, this is provided the value of 𝐿 is not equal to zero.

We want to apply this to our standard trigonometric limit result, the limit as π‘₯ approaches zero of the sin of π‘₯ divided by π‘₯ is equal to one. So we’ll set 𝑓 of π‘₯ to be the sin of π‘₯ divided by π‘₯ and 𝐿 to be equal to one. Since the value of 𝐿 is not equal to zero, we can use the fact that the limit of a reciprocal is equal to the reciprocal of a limit. This tells us the limit as π‘₯ approaches zero of the reciprocal of the sin of π‘₯ divided by π‘₯ is equal to the reciprocal of one.

Of course, the reciprocal of one is just equal to one. And we can take the reciprocal of the sin of π‘₯ divided by π‘₯ to give us π‘₯ divided by the sin of π‘₯. So we’ve shown the limit as π‘₯ approaches zero of π‘₯ divided by the sin of π‘₯ is equal to one.

Let’s see how we can use this to evaluate the limit given to us in the question. First, we’ll take the constant four outside of our limit. Next, we notice we’re taking the quotient of two squares. So by using our laws of exponents, we can instead square the entire quotient. We can now rewrite this by using the power rule for limits. Which tells us for a positive integer 𝑛, the limit as π‘₯ approaches π‘Ž of 𝑔 of π‘₯ to the 𝑛th power is equal to the limit as π‘₯ approaches π‘Ž of 𝑔 of π‘₯ all raised to the 𝑛th power. In other words, the limit of a power is equal to the power of a limit. This gives us four multiplied by the square of the limit as π‘₯ approaches zero of π‘₯ divided by the sin of five π‘₯.

Now we want to evaluate this limit by using the limit where we came up with earlier. However, we can see we’re taking the sin of five π‘₯ in our denominator. The limit where we came up with only has the sin of π‘₯ in its denominator. We’ll get around this by replacing all values of π‘₯ in this limit rule with five π‘₯. So this gives us the limit as five π‘₯ approaches zero of five π‘₯ divided by the sin of five π‘₯ is equal to one.

We have to be careful at this point since we now have the limit as five π‘₯ approaches zero instead of π‘₯ approaching zero. However, if five π‘₯ is getting closer and closer to zero, then π‘₯ is getting smaller and smaller. In fact, π‘₯ is getting closer and closer to zero. So we can just rewrite this limit as π‘₯ is approaching zero.

Next, we can take the constant five outside of our limit. In fact, we can just divide through by this constant of five. And this tells us the limit as π‘₯ approaches zero of π‘₯ divided by the sin of five π‘₯ is equal to one-fifth. So we can evaluate this limit. It’s equal to one-fifth. This gives us four multiplied by one-fifth squared. And we can calculate this expression to give us four divided by 25.

So we’ve shown the limit as π‘₯ approaches zero of four π‘₯ squared divided by the sin squared of five π‘₯ is equal to four divided by 25.

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