Luyện nói tiếng Anh bằng Shadowing qua video: The Limit (do not use L'Hospital rule)

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Okay, let's talk about the limit as theta goes to 0 of sine theta over theta.

Yes, we put 0 right here and right here, we end up with 0 over 0.

So maybe let's just go ahead and do the Lafittos rule.

Let's put on d, d theta, d, d theta.

So we are showing work, and our teacher will be happy, right?

So this is the limit as theta goes to 0.

And then when you differentiate sine theta, we get cosine theta.

And then when we differentiate theta, we just get 1.

And then now we can just put the 0 right here.

And in the end we are saying this is just cosine of zero.

Of course we can work this out.

This is just one.

That's it.

Took us about like 37 seconds.

That's it. And yeah.

Okay.

Has this ever happened to you guys?

You did this on the test.

But in the end you end up with a big X like that.

And you end up with zero points, zero marks for this question.

If this has happened to you, leave a comment down below.

And you might be wondering, what's so wrong with this?

Yes, we do the L'AbiDoc rule, and this is a 0 over 0 situation.

Is it because I didn't show work?

I forgot to put down this is a 0 over 0 situation at Lh.

Maybe I just forgot to show work.

Okay, now here's the problem.

The issue is that when you differentiate sine theta, Yes, the answer is cosine theta, but how do you know that?

Well, if you guys look back to the proof of the derivative of sine theta,

in fact, you need to first know what the limit as theta goes to 0 of sine theta over theta is first, before we can talk about the derivative of sine theta.

So, in other words, we cannot use derivative, we cannot use Lapidus' rule to figure this limit out, because otherwise, this is what we call the circular reasoning.

That's really bad, right?

Therefore, we must use another way to figure this out, and let's do it right here.

The way that we actually need to do it is by using geometry.

And first of all, let's just talk about we want to get the limit as theta goes to 0 plus.

And of course, let's put down side theta over theta.

Well, you'll see this is actually what we need, because later on we can also use this to figure out the limit as theta goes to 0 minus.

And you'll see you'll end up with the same answer, and then we can conclude the theta goes to 0 that limit.

Well, what we need to do first is, we are going to begin by looking at the unit circle.

And because we know theta is going to 0+, that means the angle is small and it's positive.

So we just need to consider the first quadrant.

So let me just draw the picture right here for you guys.

Let's say this right here is my unit circle.

And because unit circle, so we know this right here would be 1.

since the angle is small, so let me just put this down right here.

Right?

I just need a first quadrate.

That's all.

Now, I know this is one, of course.

Okay, next what we need is, let's first look at this arc.

And on the circle, we know that the arc has a formula, right?

The R-length formula.

We just need two things.

First, we need the R, the radius, and then next we need the angle theta in radians.

And of course, we are all doting out this angle theta is in radians already.

So in this case, I just need to do 1, which is the radius, times the radians theta angle right here.

So in other words, the arc length is just theta.

This represents the angle, this right here represents the length.

So that's the first thing that we need.

Next, what we'll do is, I will look at the inside and right here,

starting at this point, I'm going to draw a line straight down so that we can form a right angle.

I will also figure out a formula for this length.

And you can see that we have a right triangle now.

If you look at this right triangle, we do know the hypotenuse, and this will be the opposite side.

So let me just indicate this for you guys right here.

we have the opposite, that's the opposite, and then the hyponius is just 1.

And let me just indicate, over hyponius, and hyponius is 1.

Opposite over hyponius is what?

Yes, it's just sine, and of course we have that angle.

So therefore, this length is just sine theta.

Alright, we're making good progress.

Next, I will be looking at this point right here, which is 1 comma 0, and I will draw a line straight up.

In fact, this is also the tangent line.

You'll see tangent line.

Alright, so you see this right here, but in fact I'm not going to draw all the way, because what I will do is, I'm going to actually extend this right here, so I'll just put on some dot dot dot,

and then I will stop right here, alright, when they meet.

Well, in this case, I can create another right triangle But the problem is,

I don't know how big this is So I cannot use sign right here anymore, unfortunately, right?

But it's okay, because we know from here to here is 1

So if you just look at this bigger right triangle here This right here is once again the opposite

but this time this right here is the adjacent that we know

so just put on a d j

which in this case is also one what's opposite for adjacent yes it's tangent

so right here as told you guys earlier in fact that's the tangent line

so we get this right here equal to tangent theta like that cool Cool, huh?

Now, we have three things to compare.

The green length and then the red curve and also the blue length, right?

The blue line.

Well of course this right here is the smallest, this right here is the second, this right here is the biggest, right?

Because you see pretty much this right here is between of this and that.

From the picture you can see that's the biggest.

So let me just put this down right here for you guys.

First of all we know sine theta which is this length right here is the smallest.

Next, we have the r length, which is the theta, and then tangent theta is the biggest.

And, of course, we can put down some inequality.

It goes like this.

Well, theta changes.

In fact, theta is approaching zero plus.

So, in fact, let me also include the equal sign right here.

100

This is still true, isn't it?

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Now, don't forget what we are trying to achieve.

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This is what we have.

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We want to achieve sine theta over theta, but we only have this.

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But it seems like theta should be in the denominator.

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So we are going to take the reciprocal of all this.

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And notice, everything right here, they are all positive because we have the theta is just a positive small number, right?

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So when I take the reciprocal, I get 1 over sine theta.

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And because everything is positive, you can just flip the inequality.

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And you get this right here, and then this right here is 1 over theta.

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And then you can just flip that as well.

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If you have 1 over tangent, that's just cotangent.

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In another word, you get cosine theta over sine theta.

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Well, well, I want to have the sine theta on the top, so why don't we just multiply everybody by sine theta right here?

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And when we do that, this times this is, of course, just 1, and then, of course, sine theta is positive.

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When you multiply by a positive number, the inequality stays.

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So I'll just put down sine theta on the top over theta, and then this times that, of course, the sine theta cancel we have this right here,

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which is just cosine theta.

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So this is what we have.

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What's next?

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Well, I'm trying to figure out the limit.

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From this inequality, I'll just take the limit as theta approaching 0 plus.

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So I will do it here.

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And then I will also do it here, the limit as theta approaching 0 plus.

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And then, of course, we also do it here.

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Right?

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Cool.

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This right here, 1 is always 1.

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So this is 1.

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You maintain the inequality.

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And then, this is exactly what we're trying to find out.

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I don't know yet, so let me just write this down.

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The limit as theta approaching 0 plus sin theta over theta.

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And then, this right here is what?

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You put 0 plus right here, cos 0 plus.

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Yes, we did it over there, but this was not legit.

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You get one.

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Now, you see, this limit is in between one and one.

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Therefore, we can use the squeeze theorem to argue that.

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So I'll just put this down by the squeeze theorem.

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Yes, T stands for stiff as well.

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right?

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Anyway, we can conclude that the limit as theta approaching zero plus of sine theta over theta,

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this right here, has to be equal to 1 as well, isn't it?

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And as you can see right here, we are done, and we got this right here.

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Cool.

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Yes.

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Now, we are not done yet.

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We also have to look at the zero minus version.

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So let me just put this down right here for you guys real quick.

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If you take a look of the limit as theta approaching 0 minus of sine theta over theta.

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Well, this right here is actually not that bad at all.

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Remember, when we have 0 plus, this is as saying, you know, theta is like about 0.0001.

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When you have 0 minus, it's just a negative of that version, pretty much, right?

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So, what you can do is, and depending on how much we're going to show,

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You can just look at this as sine of 0 minus, and then on the bottom you have 0 minus like this.

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Well, sine is an odd function.

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This right here is a negative number.

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So this right here, as I say, you can just take the negative on the outside because sine is an odd function.

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So you have negative of sine, and the input now, it will be the positive version, namely 0 plus.

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Likewise, 0 minus is as a negative of the 0 plus.

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Negative, negative, cancel.

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This right here gives you 1 because of that.

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It depends on how much work you have to show, but this right here, hopefully, is a good enough argument for many of you guys.

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And now, as we know, when we have the right limit and also the left limit equal to the same number,

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we can conclude that When we have the limit as theta goes to just zero of sine theta over theta,

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this right here has to be one.

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Finally, we did it.

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This is what I call the limit.

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This right here is the limit.

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Why?

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Because it happens a lot.

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Because this right here also happens a lot.

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It hurts a lot of students.

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And once again, leave a comment down below and let us know

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if this happened to you if you did that before or not.

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And in fact, I do feel guilty because sometimes I do this myself as well but I do it in my head.

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So as long as you don't write it down, we are good.

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Anyway, hopefully you guys all like this and let me know

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if you guys have any questions and if you guys are new to my channel, be sure to subscribe.

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You guys can find a lot more interesting my videos on my channel and thank you guys so much for that.

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And as always, that's it.

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Video này có 181 câu và 2020 từ để luyện shadowing. Phần lời nói dài 12:07. Người nói nói với tốc độ tự nhiên, khoảng 167 từ mỗi phút, gần với hội thoại hằng ngày. 86% số từ nằm trong 3.000 từ tiếng Anh thông dụng nhất; phần còn lại nên xem trước khi luyện.

Từ vựng quan trọng trong video

15 từ đáng học trong video, kèm phiên âm và nghĩa:

Từ	Phiên âm	Nghĩa
angle danh từ	/ˈæŋ.ɡəl/	gốc
minus	/ˈmaɪ.nəs/	trừ
inequality danh từ	/ˌɪn.ɪˈkwɑ.lɪ.ti/	bất bình đẳng, không bình đẳng
triangle danh từ	/ˈtɹaɪˌæŋɡəl/	tam giác, hình tam giác
formula danh từ	/ˈfɔː.mjʊ.lə/	công thức
derivative danh từ	/dɪˈɹɪvətɪv/	đạo hàm
depend động từ	/dɪˈpɛnd/	phụ thuộc
cancel động từ	/ˈkæn.sl̩/	hủy, hủy bỏ
adjacent tính từ	/əˈd͡ʒeɪ.sənt/	bên cạnh, liền
squeeze động từ	/skwiːz/	đè, bóp
radius danh từ	/ˈɹeɪ.di.əs/	xương quay
theorem danh từ	/ˈθiərəm/	định lí
argue động từ	/ˈɑɹ.ɡju/	tranh luận
compare động từ	/kəmˈpɛɚ/	so sánh, so
extend động từ	/ɛkˈstɛnd/	kéo dài

Phát âm cần chú ý

Người nói dùng 24 dạng rút gọn, ví dụ don't, I'll, I'm. Hãy nói theo dạng ngắn đúng như bạn nghe.

Âm “th”: theorem /ˈθiərəm/, theta /ˈθiːtə/
Từ dài — đặt trọng âm cho đúng: inequality /ˌɪn.ɪˈkwɑ.lɪ.ti/, derivative /dɪˈɹɪvətɪv/, differentiate /dɪf.əˈɹɛn.ʃi.eɪt/, geometry /d͡ʒiˈɑ.mə.tɹi/, reciprocal /ɹɪˈsɪpɹək(ə)l/

Cách luyện với video này

Nghe hết video một lần, chưa cần nói, và ghi lại những từ bạn chưa biết.
Bắt đầu ở tốc độ 0,75×, nói đuổi từng câu, rồi quay lại tốc độ bình thường khi đã quen.
Ghi âm giọng mình rồi so với bản gốc, chú ý các từ như angle, minus, inequality.

Phương Pháp Shadowing Là Gì?

Shadowing là kỹ thuật học ngôn ngữ có cơ sở khoa học, ban đầu được phát triển cho chương trình đào tạo phiên dịch viên chuyên nghiệp và được phổ biến rộng rãi bởi nhà đa ngôn ngữ học Dr. Alexander Arguelles. Nguyên lý cốt lõi đơn giản nhưng cực kỳ hiệu quả: bạn nghe tiếng Anh của người bản xứ và lặp lại to ngay lập tức — như một "cái bóng" (shadow) đuổi theo người nói với độ trễ chỉ 1–2 giây. Khác với luyện ngữ pháp hay học từ vựng bị động, Shadowing buộc não bộ và cơ miệng phải đồng thời xử lý và tái tạo ngôn ngữ thực tế. Các nghiên cứu khoa học xác nhận phương pháp này cải thiện đáng kể phát âm, ngữ điệu, nhịp điệu, nối âm, kỹ năng nghe và độ lưu loát khi nói — đặc biệt hiệu quả cho người luyện IELTS Speaking và muốn giao tiếp tiếng Anh tự nhiên như người bản ngữ.

Phương pháp shadowing: đọc hướng dẫn từng bước đầy đủ →

Luyện nói tiếng Anh bằng Shadowing qua video: The Limit (do not use L'Hospital rule)

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Phương Pháp Shadowing Là Gì?