Limits of Trigonometric Functions

This calculus video tutorial provides a basic introduction into evaluating limits of trigonometric functions such as sin, cos, and tan. It contains plenty of examples and practice problems.
Introduction to Limits: • Calculus 1 - Introduct...
How To Evaluate Limits From a Graph:
• How To Evaluate Limits...
Evaluating Limits By Factoring:
• Evaluating Limits By F...
Limits of Rational Functions :
• Limits of Rational Fun...
Limits of Radical Functions:
• How To Evaluate Limits...
Limits of Trigonometric Functions:
• Limits of Trigonometri...
___________________________________
How To Find The Limit at Infinity:
• How To Find The Limit ...
Infinite Limits + Vertical Asymptotes:
• Infinite Limits and Ve...
Limits at Infinity With Radicals:
• Limits at Infinity Wit...
Limits of Absolute Value Functions:
• Limits and Absolute Value
Limits of Composite Functions:
• How To Find The Limit ...
The Squeeze Theorem:
• Squeeze Theorem
____________________________________
Limits and Logarithms:
• Limits and Logarithms
Limits of Exponential Functions:
• Limits of Exponential ...
Piecewise Functions - Limits:
• Piecewise Functions - ...
3 Step Continuity Test:
• 3 Step Continuity Test...
Continuity and Differentiability:
• Continuity and Differe...
Limits - Test Review:
• One Sided Limits, Grap...
_____________________________________
Final Exams and Video Playlists:
www.video-tuto...
Full-Length Videos and Worksheets:
/ collections

[@TheOrganicChemistryTutor]

Final Exams and Video Playlists: www.video-tutor.net/ Full-Length Videos & Worksheets: www.patreon.com/MathScienceTutor/collections Next Video: kzbin.info/www/bejne/rqepoqKjhNV_qNU

[@MikeChipili]

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[@mathias_dangi_265]

Are you able to prove that (3w/sin3w)=1, because the only thing I could remember in my calculus class is to prove that (sinx/x)=1. Or do you take a resprocal of it?

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How do you do that

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[@wenwen8237]

can anyone answer me why it should be radian mode btw

[@jemilieredito2498]

(sin x) (x) = x sin x (sin x) (sin x) = sin^2 x So it should be limit of sinx/x as x approaches 0 times limit of sin x as x approaches 0. Though the answer is still 0, there’s a problem in the process. Thanks.

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I thought I'd be the first to correct this but here it is

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This correction is for what time in the video?

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This man blessed me before my limits exams

[@husseinalhadi5471]

Excellent video at 13:43 mark shouldn't Sin ^2 = sin (x) X Sin (x) not sin (x) X x

[@agent-if9jb]

agreed.

[@marknzuri9119]

Yeah I thought so too!

[@aliu760]

yeah I saw that mistake

[@benjaminm3180]

Since sin(x) using the unit circle equates to zero, and x is approaching zero (so you plug in zero), either way you write it will be multiplying by zero. I’ll admit I had to pause the video to think about it for a minute to completely understand what he meant but the way he wrote it is still technically viable.

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[@laizat.1300]

Isn't it that sin^2 x = (sin x)(sin x)? Why is it that at around 13:36 of this video, the sin^2 x became (sin x)(x)? Although, the final answer is still gonna be the same. I just got confused. I am hoping for your reply. Thank you. :)

[@m.rizaarrafi411]

i was about to say that, i think he made a little mistake right there

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[@ninjanigabiyown4253]

The problem where there is sin²x It should be (sinx)(sinx) instead of (sinx)(x).

[@evansraila5519]

It'll still be the same i think...

[@Bbqeee]

​@@evansraila5519 yes bc (sinx/x) will be 1 And as x→0 (sinx) will be 0 So 1*0 will be 0

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[@ianruiz6350]

I'm just here scrolling down through comments seeing everybody somehow understanding even though he never explained the why of things. 3:14 how did he know that he had to multiply by 5? 5:39 how did he know he had to multiply top and bottom by 2? 7:33 how did he know he had to do THAT? I just don't see a pattern at all. It feels like he's explaining how to do the specific problems rather than teaching us how to analyze problems on our own. I'm gonna be solving all of these thinking I understand them then when the tests arrives I won't even be able to know where to start or how to tackle the problem.

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you guys learn these at college ? i'm in 11th grade and we have these stuff

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Trying to find the limits of trigonometric functions are probably one of the most complicated things when entering to the realm of calculus.

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[@georgesadler7830]

MR. Organic Chemistry Tutor, thank you for calculating Limits of Trigonometric/Transcendental Functions in Calculus. Thanks to the viewer(s) for correcting the error on (Sinx)^2/x in the video. Although the final answer is correct, the method is incorrect.

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[@charliechaplin53]

i think there was a mistake instead of sinx*sinx, you wrote sinx*x. Infact limit when x approaches 0 of sin^2(x)/x = 0

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[@skog3372]

I might be wrong but isn't sin^2(x) = sin(x) * sin(x)? he says it equals sin(x) * x

[@calista.monteiro]

Actually

[@IMathTutorial]

Thank you

[@kpopscenario1406]

guys u don’t need radian mode for it and easier way would be just writing pie as 180 and you’ll get it all. for example pie/2 is 90 degrees, u can simply just do 180/2 and you’ll get 90 or another example 2pie it’ll be 180 x 2 which would be 360 degrees or a harder example 3pie/2 u would simply do 3 x 180 which would be 540 then divide it by 2 and you’ll get 270. thank me later!

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I came here because my professor needed me to evaluate limits of trig functions involving ln but without a calculator on an exam

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[@charliebardey1990]

can someone explain why at 13:48 (sin^2 x)/x becomes (sinx)/x * x???

[@scienceappreciator]

typo

[@emilycallicott7475]

Hi, not sure if youre still answering questions on this vid, but at 13:35, you said how sin^2(x) was the same as (sin(x))^2 but when written out, you put Sin(x) times X instead of Sin(x) times Sin(x). Am I thinking of it wrong?

[@JKMath]

That was a mistake on his part. You are correct! It should have been sin(x) times sin(x), not sin(x) times x.

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[@kamrulhaquealmi2693]

if you're looking for why to use radian? then here's an explanation - In radian we use number, like the pie itself, π=3.1416... in degree which is equivalent to π radians=180° So, if you look into the function like sinx/x, in here x=some number; because lim(x→0) not x→0°. x does not mean degree of any angle...that's why we have to use radian.

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Me literally crying because i can't use a calculator in my final

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[@ashleyfontanilla2262]

There is a better way to approach the problem presented at 6:00, in case if anyone is confused by the arbitrary, "just switch the 5 and the 2." Instead of "just switching," multiply the numerator and the denominator by 2/5 to maintian a logical approach. In the denominator the 5's cancel out leaving a 2x, and just keep the 2/5 that pertains to the numerator on the outside as it was exemplified in the previous problem.

[@scarleteyes415]

Very easy way to do it: Before multiplying anything, we have the fraction "sin(2x)/5x". Since the fraction is already divided by 5, we take 1/5 outside the limit, so the fraction becomes: 1/5 (lim sin2x/x). Now we just solve it like we did in the previous question, the only difference is that we have 1/5 outside the limit. We multiplay the equation by 2/2 and we get: 1/5 (lim 2sin2x/2x). Now we just take out the 2 in the numerator outside the limit the same as the previous question, hence it's multiplied by 1/5 that is also outside the limit: 2/5 (lim sin2x/2x). So, we put Y = 2x and we get: 2/5 (lim siny/y) as y goes to zero, so the latter term is equal to 1, therefore 2/5 (1) = 2/5

[@heavysaber9431]

I agree, his methods are somewhat unorthodox to a logical approach to math, but it still works anyhow. Still, its better to have general logic in tackling these problems instead of just "switch these numbers" without knowing the reason why or how the value within the function should still remain the same while arranging the numbers, manipulating numbers isn't that simple.

[@miguelconfesor4701]

The 2 and 5 can be switched because we are dealing with multiplication.

[@MoguMogu818]

Honestly great I can pause and learn from this. but at the same time I wish I wasn't so goddamn stupid that I can't seem to understand calc 1. Idk how I was able to go through math classes in high school and get straight A's each time and then suddenly freeze up when the word limit comes up. Man I am dumb. But hey that's why I'm studying rn.

[@titananni3]

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[@mariamhaytham2482]

in the example of sin^2x/x why did we multiply sinx/x by X and not sinx while noticing that he already said that it is supposed to be sinX multiply sinx/

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this genuinely saved me thank you so much

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Thx for ur video soooo much cuz we really get a terrible textbook for calculus QAQ

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[@reenweru9416]

I had given up on this until you said "make sure the calculator is on radian mode"

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Thank you so much....You're such a great teacher by the way😊🙏🙏

[@user-mh5oq4vl4d]

Great thanks to the voice behind the video

[@julemarielumapas3586]

I don't know if it's just me but on the solution on 2:00 the 1-cos(0.01) divided by 0.01 is equal to .005. But no matter how I solve it in my calculator the anwer is always 4.99995. Why is it like this?

[@nadanassar]

Me too but I don't why?

[@mattmo8648]

at around 6:50, you change around the lim to y instead of x, creating y=2x. why not just stay with sin2x/2x, and make it equal to 1? what is the reason for switching over to the y based equation?

[@mattmo8648]

side question - where do you pull the 4/4 from at 11:27? i understand the tan 4x, but are we not multiplying by 3/3 because 3x got moved? what about x/x? the last problem made me think we had to use each independent variable. i'm a little confused on that.

[@JKMath]

The reason he switches to y is to show you that it is still a form of the same special limit. That step is not completely necessary in your work, but he does it anyway to remind you that it is a special limit that is equal to 1. So in the end, all you need to know is if you have the limit as x approaches zero of sin(x)/x where as long as whatever is inside sin is also in the denominator, the limit is equal to 1. Hope that clears things up for you!

[@JKMath]

@@mattmo8648 For your side question, what you need to know is that you are always allowed to multiply by a form of 1. Anything divided by itself is 1, and multiplying anything by 1 does not change that value. So, since 4/4 is just equal to 1, we can multiply it by the limit and that allows for the use of the special trig limit. We are not multiply by 3/3, instead the 3x in the denominator was just moved out so that it is easier to see where the special limit is located. So, when determining what form of one to multiply by, look inside of your trig function (in this case sin(4x) ), and whatever is inside, you want that to be in the denominator. That's why you multiply by 4/4, because then there will be a 4 in the denominator that is multiplied by x that can then be moved under sin(4x) to then create a form of the special limit. I hope this makes sense, if not (this can be difficult to describe with just words), feel free to check out my own tutorial video on this topic that is on my channel, I'll link it here for you. I have all kinds of calculus tutorials that you may find helpful! Here's the link: kzbin.info/www/bejne/gaDHnql6rqyNpqs

[@forthehomies7043]

Dare I say you make calculus... FUN !

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[@da_knug]

Great video, thank you!

[@HypedRogue]

Loved this video it really helped me catch up in my calc classes- one problem near the end though, I'm pretty sure sin^2(x)/x does not equal sinx/x*x, it equals sinx/x*sinx, although the limit still comes out the same.

[@gaiga494]

x*sinx/x you mean, but yeah limit comes out the same since sin(0)=0 and 0=0

[@kunshanand1433]

incredibly helpful

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I love you thank you you saved my life

[@graphetes]

it would be get if you didn't use l'hospital rule because most calc 1 classes require you NOT to use that rule. Either way I love all your videos and thanks for the help in all categories.

[@nomnom756]

wait so is there no way to determine the limit of a trig function without plugging in the value>