Limits of Trigonometric Functions
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- เผยแพร่เมื่อ 25 ก.ย. 2024
- 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
Limits - Free Formula Sheet:
bit.ly/3T3dD2X
Limits - Free Formula Sheet: bit.ly/3T3dD2X
Final Exams and Video Playlists: www.video-tutor.net/
Full-Length Videos & Worksheets: www.patreon.com/MathScienceTutor/collections
Next Video: th-cam.com/video/xvFqomOpLrs/w-d-xo.html
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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
@@wm396 depends what calculator you have?
ur comment is a life saver
can anyone answer me why it should be radian mode btw
(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.
Thought i was the only one. (^^)
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I thought I'd be the first to correct this but here it is
This correction is for what time in the video?
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This man blessed me before my limits exams
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Excellent video at 13:43 mark shouldn't Sin ^2 = sin (x) X Sin (x) not sin (x) X x
agreed.
Yeah I thought so too!
yeah I saw that mistake
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.
Yeah! Sure
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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. :)
i was about to say that, i think he made a little mistake right there
Yup, it's a mistake
i thought i was retarted
Yes
Looking at the comments checking if I'm the only one. 😆
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The problem where there is sin²x
It should be (sinx)(sinx) instead of (sinx)(x).
It'll still be the same i think...
@@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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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.
up
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can someone explain why at 13:48 (sin^2 x)/x becomes (sinx)/x * x???
typo
this video was very helpful for my online classes in calculous. Its hard to learn math through just pdfs. Glad I saw this!
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this is actually such a straightforward and simple explanation thank you very much for making this video it saved me in AP calc since I was trash at trigonometry.
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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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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I had given up on this until you said "make sure the calculator is on radian mode"
Simplicity is the Key 🗝️🗝️. Love this Topic
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 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
I finally understand how to do this after struggling with it for 3 years... thanks dude
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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?
That was a mistake on his part. You are correct! It should have been sin(x) times sin(x), not sin(x) times x.
@@JKMath Omg thank you! I thought I was going insane lol!
I might be wrong but isn't sin^2(x) = sin(x) * sin(x)? he says it equals sin(x) * x
Actually
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you guys learn these at college ? i'm in 11th grade and we have these stuff
Yup they also have algebra and geometry and some colleges have more basic ones. A lot of technical universities offer calculus but that’s like basic
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hmmmm
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?
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.
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!
@@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: th-cam.com/video/KodmvDyGZpI/w-d-xo.html
Absolute king - the greatest tutor on planet earth
I found the sin7x/sin3x strategy very helpful.
The first opening sentence. Thank my dude everything make sense now.
Hey thank you for this awesome explanation you just saved my time
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?
Me too but I don't why?
you're doing gods work thank you
thank you king u carried me in physics last year and ap calc this year
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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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.
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/
Thank you so much....You're such a great teacher by the way😊🙏🙏
What if sinx isn't divided by x? My questions are along the line of 'Find lim (x->0) -2sin3x'. I can't find the answers anywhere and am taking my course online...
Your question is the limit as x approaches 0 of -2sin3x, right? Direct substitution is enough since the limit is linear. So it will be -2sin(3(0)) which will be just 0. For fractions I think the rule is enough
Dare I say you make calculus... FUN !
Thank you so much I owe you my life
What textbook do you think I can use to get more help in this?
wait so is there no way to determine the limit of a trig function without plugging in the value>
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.
Great thanks to the voice behind the video
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.
this genuinely saved me thank you so much
I have a question:at 9:10,why is the limit as w goes to 0 of w over sin(w) equal to 1?
w / sin w is 1 because it is the inverse of sin w / w:
w / sin w = 1 / (sin w / w)
w / sin w = 1 / 1
w / sin w = 1
Very clear and easy to understand. Thank you.
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.
x*sinx/x you mean, but yeah limit comes out the same since sin(0)=0 and 0=0
this man's a genius.