A sine limit It can be shown that 1−x^2/ 6 ≤ sin x/ x ≤1, for ...

Question

A sine limit It can be shown that 1−x^2/ 6 ≤ sin x/ x ≤1, for x near 0.

Use these inequalities to evaluate lim x→0 sin x/ x.

Accepted Answer

Hello there today we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Assume the inequality one minus W squared divided by 12 is less than and equal to sin of W divided by W is less than or equal to one for W near zero holds use this to evaluate the limit as W approaches zero of sine of W divided by W OK. So we're asked for this particular prom, we're trying to use our inequality that's provided to us and we're trying to use this inequality to help us evaluate the specific limit. So now that we know that we're ultimately taking our inequality and we're using it to help us evaluate our specific limit. Let's read off our multiple choice answers to see what our final answer might be. A is zero, B is negative one C is one and D is infinity. OK. So first off, in order to help us solve this problem, let us define the known variables and the target variable. So we need to figure out what variables are known to us or variable is known to us. And we need to determine what the target variable is. So we know that our known variable is W and our target variable in this case is going to be our limit, which is evaluated as W approaches zero. So the limit as W approaches zero of sine of W divided by W, so that is our target variable is this limit. So now we need to use our given inequality. So our given inequality once again is one minus W squared divided by 12, which is less than, or equal to sine of W divided by W which is less than or equal to one. And we're gonna be using this inequality to evaluate the limit as W approaches zero. So let's make a note here. So this is when A W approaches zero fantastic. So as W approaches zero, so let's make this another note since this is very important. So as W approaches zero, we need to note that it's very important. Let's do a little and symbol, it's very important to note that one minus W squared divided by 12 will also approach one. So one minus W squared divided by 12 will also approach one. So keeping this in mind our next step is we need to recall and apply the squeeze theorem. So using the squeeze theorem as we should recall and note that sense, one minus W squared divided by 12 is less than, or equal to the sine of W divided by W which is also less than, or equal to one. And both, we need to note, let's make another answer. I need to note that one minus W squared divided by 12 and one will approach one. As W approaches zero. We can conclude that we can ultimately conclude that the limit which is as W approaches zero of sine of W divided by W will also equal one. And that is our final answer. Hooray, we did it. So this problem is very easy to solve. As long as you understand, all the background information that is required to make these assumptions and make these important note like, you know, notices that we did. So you need to be able to recall and use squeeze theorem and you need to be able to evaluate limits. And if anything I said doesn't make any sense, I would say just go read up on your background theory to get a better understanding of this type of problem. And once you do solving a problem like this is very simplistic and very easy. So with that said, our final answer without a doubt has to be the letter C which is one. Thank you so much for watching. Hopefully, that helped and I can't wait to see you in the next video. Bye.

A sine limit It can be shown that 1−x^2/ 6 ≤ sin x/ x ≤1, for x near 0.
Use these inequalities to evaluate lim x→0 sin x/ x.

Key Concepts

Squeeze Theorem

Limits

Trigonometric Functions

Watch next