Calculating a Limit Involving sin(x)/x as x approaches zero

[worldismine3]

what happens if you have lim x to 0 sin(x^3)/x?

[zMagickz]

you deserve my tuition

[ShadowDog415]

wow you're left handed so am I! lols

[musicISlyf1]

why does it become cos?

[battousai12345]

@musicISlyf1

Because tangent is equal to sin/cos or (Sin/1)(1/cos)

[xxo0O77]

again thank you so much! my teacher moves soo fast because it is an calculus class for science majors, so much love!

[villex4]

we have a question like that @ 3:25....my teacher wrote wrong solution....

[MrLikenoone]

can someone tell me lim sin(x)/x when x goes to 0 . x in degrees ? why ?

[MindLabOne]

@MrLikenoone π/180.

[MrLikenoone]

@MindLabOne yes of course

[killaplaya222]

are you a wizard?

[hatemailism]

Genius 

[snoogygaming]

"Useful limit to know, it is just another random little formulas worth memorizing"

LooooL XD

[MultiMijin]

isnt the answer for the last example 1?

[bounasif]

your second example was soo confusing!

[redrum41987]

who's crammin for exams?

[AliveDog]

@redrum41987 calc exam tomorrow. yep yep

[IHateSchoolPeople]

so, what if it was approaching infinity?

[prasiprassi]

i hope you get paid for subscribe and tnx for the video

[kayongoedward]

thnxxxxxxxxx *on my knees*

[gangliums]

wait you did division by zero :S?



[litcy02]

I have a question, but why didnt you multiply 6t to 1/cos(6t) and get 6t/cos(6t)?

[Ssendecki]

l'Hospital here is rather funny because the very same limit is used when evaluating the derivative of sin(x)

[Lateilla]

wat about if you have a problem like (x * (cos3x/sin3x)) then wat? sure the top and bottom both have 3x but wat do i do with the cos? i'm nt really sure where to go with this problem.

[Ssendecki]

@Lateilla cos0=1

[Sk8r277]

0 divided by 0 is undefined... i don't get it.

[GhostSarge]

@Sk8r277 You cannot divide by zero; It just doesn't work; It's not possible; It's like a giant limit.

It's indeterminate/undefined.

[thenoobersaur]

I didnt know that sin(x)/x = 1 was an identity. i was stuck on a problem for sooo long.

i dont usually leave comments. but i only leave them to the ones that deserve one. thanks!!!

[adnshr]

thank you very much

great teacher

[patrickJMT]

@adnshr thanks : )

[sciencennerd16]

What about lim x-- 0 xsinx/1-cosx

[tweedyasianbird]

i was struggling with this for a long time and i grew frustrated for not understanding it. then by 2:20 of your lesson i finally understood. thank you so much, i greatly appreciate it!

[patrickJMT]

@tweedyasianbird you are very welcome

[jojogirl2008]

Thanks So Very Much!!

[cteeput]

Thanks a lot!

[karanaussie]

you make calculus very easy...keep up the good work. cheers from india.

[KapustaCuber]

I'd figured this out before I even clicked on it using L'Hopital's rule. I'm only a Sophomore in High School, haven't reached Algebra II yet. 2/3rds of my informal calculus lessons have come from this channel, thank you very much.

[dbogmbinj]

2t/sin2t is one. I don't understand?

According to formula sin2t/2t is one but not the other way around? How did you got this. By the way thanks for all the videos, you been of great assistance. I have exam 2 days from now and studying for couple of days with your videos. Greetings from Serbia.

[bhasder998]

@dbogmbinj i'm also confused about it....

[1990GIGA]

its amazing. vau

[AnimangaArtist]

That was like...magic. Your explanation...simply beautiful..

[youwillbemywife]

ok we know that the limit of sin(x)/x=1 ( could we use this as an identity?), second question,what if my instructor tells me to prove that sin(x)/x=sin(n)/nx? btw, i love your videos, it helps a lot of my callculus I class

[dajakesta1234]

but if u use lh then wut would the theta be?

[Soccerjunky04]

For fear of sounding incredibly slow I dare not ask, but I have a my first Calc test tomorrow, and I don't know the difference between limits equaling D.N.E. and infinity( is there a difference?) By the way your videos have made the class so much better!

[caroleabumoussa]

@Soccerjunky04 I would like to know the answer, too!

[Topwatch288]

@Soccerjunky04 When the limit is infinity, the limit doesn't exist (DNE) because a limit, by definition, is a constant. On a test, you'd want to write DNE after showing that the limit is infinity.

[rsvitale]

@credico assuming you mean as x-->0: multiply by 9x/9x: (1-cos4x)*9x/((9x^2)*sin9x), multiply by 16/16 and seperate into parts that you know the limit of: (16/9)*((1-cos4x)/((4x)^2))*(9­x/sin9x). as x goes to 0: 16/9 -->16/9, (1-cos4x)/(4x)^2-->1/2, and 9x/sin9x-->1. (16/9)(1/2)(1)= 8/9. The limit as x goes to 0 is 8/9. hope this helps.

[moosiau]

why does lim t->0 become lim theta->0 ??

Is it a mistake or is it done for a reason?

[moosiau]

@moosiau sorry! just saw that later in the video you correct the mistake ^.^ hehe thanks

[18Zac1992]

great video! Thanks for all your help!

just a quick question, does anyone know if its possible to apply this to functions of 2 variables. say i have a function sin(x^2)+(y^2))/[(x^2)+(y^2)] and im trying to find the limit as (x,y)--->(0,0) can i make a simple substitution say u=x^2+y^2 and re-write my original function as lim as u-->0 sin(u)/u WHICH IS EQUAL TO 1???

any help would be great thanks! :D

[tutiha]

can you prove why lim sinx/x = 1 = lim x/sinx , x approaches 0? Or it is something you can just use? Thanks.

[patrickJMT]

@tutiha the proof is not trivial, but it should be in any calculus book. i do not think you would be expected to prove it in most any calc 1 class

[rsvitale]

@tutiha There's a simple way to prove it if you know how to expand sin(x). sin(x)=x-(x^3)/3!+(x^5)/5!-(x^­7)/7!+.... So sin(x)/x=1-(x^2)/3!+(x^4)/5!-(­x^6)/7!+....The series goes on infinitely if you want an exact equation. As x goes to zero you can see that all of the terms except for 1 go to zero as well, so lim sin(x)/x, x->0 =1. I'm not sure if you've seen taylor expansion or not but maybe this helps. sorry if it doesn't.

[patrickJMT]

@rsvitale well, at this point, to derive the taylor series expanion, you need derivatives. to get the derivative of sine, the result in this video is needed - so what you are doing is in a way circular reasoning

[rsvitale]

@patrickJMT Yeah i wasn't sure what the poster knew. I had heard of writing functions as polynomials before i knew derivatives, so if the poster had seen the series definition of sin(x) for some reason i was just giving a quick way to see that the limit is 1. But yeah it's circular if you are thinking about what a taylor series actually is instead of just knowing the polynomial for sin(x) as a definition.

[rsvitale]

@patrickJMT Oh and by the way I like your videos!

[Topwatch288]

@tutiha I think you could just rewrite lim x/sinx as lim [1 / (sinx/x)]. Then you could take the limit of both numerator and denominator, and you'll get 1.

[vampiracy]

@tutiha

I would say just think about the value of 1. If the sinx/x is going to 1, then sinx and x are going to the same thing, to make the division come out to 1. If we flip the function, we're still dividing a number by itself, so we still get 1. Think 6/6; it equals 1 as it is, and when you flip it, t becomes 6/6 again which still equals 1. Another way to think about it is raising both sides to the -1 power, or flipping both sides. sinx/x=1/1, x/sinx=1/1. Hope this helps.

[82sledge]

@tutiha : The proof is trivial: L'Hospital rule says that if a limit funcion leads to the expression like "0/0" you can replace the functions (upper and lower in the fraction) by their derivation. Derivation of x is 1, derivation of sin (x) around x=0 is also 1. Then: Lim sinx/x =1/1=1=anything else what is equal to 1 :o)

L'Hospital rule: at en.wikipedia try to find L'Hôpital's_rule

[Zhaggysfaction]

I just had a maths exam with one question about the function sin(x)/x but we had to prove it's decreasing when x is ]0,(pi/2)[

[deszczyn]

5:13 - 5:20 you say sin(6t)/6t is 1 when you substitute 0, how is 0/0 = 1 ?

[Zhaggysfaction]

@deszczyn

both the sin(6t) closes 0 and 6t closes 0 but they are never equal to 0 so it's never 0/0. it's the limit that goes to zero

[Rafi1994]

hey dude, i thing, ur vids are great, jsut letting you know though for that Tan/Sin problem, the same rules tat apply for sin apply for tangent also, only when x approaches 0, so u didint have to do all that work...tan6t/sin2t when t is approaching 0, is 3 but u dont gota do all that work

[wen609]

okay this is a miracle. the two topic my professors' test is gonna be on are : limits involving sin, and 2. applying the squeeze theorem. you going by the larson edwards book by any chance? <3

[patrickJMT]

@wen609 not larson, but stewart

[sciences8]

@patrickJMT

Thanks for you help. My teacher is very ambiguous and expects us to know everything for some reason...

[thefluffylizarazo]

u are a beast....and thanks to this, i may jst pass my midterm tomorrow....thanks!!!!!!!!!!!!

[woody189]

I love you Patrick...thanks

[smilecolorsan]

you are honestly great at this. thanks a lot.

[miladini1]

Hello,

I am not sure if you can prove this identity using L'Hopital's rule. Let me just ask you a question and you will know why you can't use that rule! can you prove to me that the derivative of sine function is cosine? how do you prove it? (Hint: you have to USE the very same identity!)

That is what I was just wondering about today (I wanted to know what is the proof for the derivative of sin(x) and then I realized that I do not know the proof for sinx/x identity until i figured it out

[dpeach06]

Another awesome video!!! Thanks so much, I love your strait-forward and easy to understand explanations. Time to ace that Calc test!

[darryllobo14]

does this also apply for a f(x,y) where (x,y) approaches zero for a function like this:

[sin(x-y)]/(x + y) as (x,y) approaches zero.... does this work, and x and y at the denominator are both absolute values, i dont think youtube allows me to put absolute signs

[darryllobo14]

@darryllobo14

and does this limit 1?

[darryllobo14]

does this also work if the limit of sin(x-y)/(x + y) as (x,y) ---> (0,0) ??

the denominator is abs x and abs y by the way

[tatromain]

Thank you!

[tatromain]

hank you!

[bluismoo]

You have a great way of reminding first year calculus students of the important algebra and trig concepts they need to know while going through the calc problems. This way, while watching the videos were not like "hold on there! how did that happen!?". You do an amazing job! If you come out with a dvd set, I will definitely buy them for super-inflated costs!

[patrickJMT]

@bluismoo i am currently in the process of getting all the videos organized; they will be available (for inflated costs!) for download in the next couple of weeks.

[123valid]

@patrickJMT Hey does that mean your current videos both on youtube and on your website will have a cost? or will they continue to be free?

BTW your a great teacher!!!! please keep your videos cost free!!!!!

[patrickJMT]

@123valid everything can be continued to be watched on both youtube and my website for FREE (not gonna change this!!) - some people wanted downloads though and this will not be free (it is not free for me to set it up and do it all!)

[123valid]

@patrickJMT Pheww...felt like I dodged a BULLET!!!! :D:D:D lol...

THANKS FOR BEING A GREATTTTTTTTTTTTTTTTT TEACHER!!!!!!

[patrickJMT]

@123valid ha, no worries : ) the reason i did made these videos was to give out some free content to people. so long as google/youtube does not charge me to upload/host videos, i will not charge anyone else : )

[Xytos]

@patrickJMT have u been to mit?

[patrickJMT]

@Xytos nope

[azala88]

L'Hospitals rule is safer and less likely to make careless mistake imo, nevertheless nice video

[patrickJMT]

@azala88 once again though, at this point in most calculus courses when people see this limit, they have not even discussed derivatives, so how could one use L'Hospital's rule?

[bamalamalamalamalama]

Your videos are very helpful, I am a first year university student who has never done calc before and I have a good prof but the extra help is much appreciated.

[wendyday]

You helped my baby out with his calc hw so I "thumbs upped" your video.

[patrickJMT]

@wendyday you have one smart baby

[t0r4g0wn4g3]

You helped me with my calc homework so I am now subscribing.

[nutella871]

So sin(x)/x = 1 = x/sin(x) ? Either one just equals 1?

[patrickJMT]

@nutella871 no, sinx / x does not equal x / sin x but they do have the same limit as x --> 0 (why is that?)

[nutella871]

@patrickJMT lol but that's the formula you have at 0:53. Hm, kind of like how x/x and 1 do not necessarily equal because on a graph, the x/x would not have a value? But in limits, that doesn't matter, right?

[patrickJMT]

@nutella871 no, that is not what i have at 0:53. what i have involves limits.

[nutella871]

@patrickJMT Ah. You're right. I see where I was wrong, thank you!

[augustineidibia1987]

Thanks

[tonyqhub5]

i've got a question... what if theta is approaching infinity? does the limit go to zero then? :\\

[cruisinces14]

sinx / x is the same with x/sinx??

[MrRabbit87]

I appreciate your videos, you and khan academy are the #1 reason I am getting through calculus with an A+.

[nuhudabomb]

in the last ones u did cant u use l'hospitals rule when its in indeterminate form

[chaoticmess2008]

Can I ask..... You said at the end that the 1/cos(6t) will just be 1/1. Don't you have to do the same thing with cos as you do with sin though, and have the same number in the numerator as you have next to the cos? (ie 6t) ????

These are all a bit confusing to me though! Your video really helped but I got a bit lost right at the end!!

[scgilleland]

@chaoticmess2008 that same sin rule doesn't apply to cos(6t), he simply plugged 0 into that part to get 1/cos(6(0)), and if you put that in your calculator you will actually get 1, cos(0) = 1, no special rules there

[roflcopter2225]

there is a really nice geometric arguement using the squeeze theorem that proves this also, but great job.

[roflcopter2225]

there is a really nice geometric arguement using the squeeze theorem that proves this also, but great job.

[roflcopter2225]

there is a really nice geometric arguement using the squeeze theorem that proves this also, but great job.

[jamskiclovek]

At the last limit you could use 2t from 6t and you would end up multiplying with only 3 so you can do 1 step less :D

Anyway thank you a lot!!! :D

[darkwolf664]

YES! You have the book I'm using!

[aparrotslife]

amazing, wonderful teacher...thanks.

[JBSauce]

I was stuck on this on the homework...thanks patrickJMT!

[ianharwood]

dude thanks. you may single handedly save me this semester lmao

[karlkarlkarl1234]

With the second example, how can:

sin(3 . 0) / (3 . 0) = 1

?

Wouldn't it equal 0/0 instead of 1 like the identity says?

[F1Racer08]

well according to the L'hospital's rule the you would get

(cos(3x)*3)/1

This would give u 3 as ur answer

[PuchTaa]

THETA IS MEASURED IN RADIANS AND VALUE OF THETA IS VERY LOW SO ACCORDING THE BOOK I HAVE AT CERTAIN TIME YOU WILL GET 1

lim sinY

y-->0 Y

Will be one

[meryn16]

Thanks again Patrick!

[pokiri446]

if i have heard this 8 years before i would have been an intelligent person.

[nhesche777]

you are a great man

[sullivanseven]

Patrick I love you brother

[Riry17]

hi, i can't solve this

lim (1-x) tan(Л/2) x ,x approaches 1

i wonder if u can help

[Riry17]

hi, i can't solve this

lim (1-x) tan (Л/2)x ,when x approaches 1

i wonder if u can help

[dual85]

I have to say. I really enjoy the way you teach. Its very calming. I rely on these videos all the time with every new section.

[protein22]

I didn't understand why 1/cos6t=1/1

[night5life]

this is because the "t" is approaching to 0.

we have 1/cost6t right? well since t is approaching 0, we just plug in 0 for the "t" value so it becomes 1/cos(6 * 0) = 1/cos(0) and looking at a cos graph or plugging into the calculator, we see that cos(0) = 1.

Therefore, 1/cos(6t) = 1/cos(0) = 1/1 = 1

[joshjm7]

What about when there is sin^2(x) on top and x^2 on the bottom

[karlkarlkarl1234]

@ joshjm7

For that, you would use log wouldn't you?

[Slayermagic]

why cant you teach at University of Toronto!!!! Aghh

[autumnofmyheart]

Thank you so much. your way of solving this is so much easier than the way I used to do it. You're the hero... LOVE!!!

[y2kx3]

ty, better than the textbook/prof by far

[olichkaluv]

AMAZING!!

[night5life]

nice!

you > my calc prof / textbook

[SaudiCanadianBoy]

what if x approahces pie over 2 ?? and f(x)= sin ( 0-pie/2) all over 0- pie/2 ??

[maths486]

same thing. limit is still 1.

[Najrani87]

thank you so much , I already got it ,but still I appreciate it :) thnx

[TWMA6]

Just what I needed to know. Thanks!

[jackiegundani]

Thank you so much! your videos have helped me a great deal

[sugarhimonkey]

Thank you so much! I'm actually going to be able to do my Calculus homework now!

[45daysinrome]

You're such a great teacher!

[andy120692]

You should get paid by youtube for these lessons

[GoldXpert]

yeah definately. And i think he does since he's a youtube partner.

[EmpressRan]

<3333333333333333333333333 x10 ^ infinity

[navalarch07]

you are a life saver

[munak75]

dude, you have no idea how helpful this is to me..thanks a lot!!

:D

[therotaryrocket]

Dude I've been watching all types of math videos on youtube, yours are some of my favorites. They definitely help me out. Thanks!

[TheeBycth]

Thank you so much! I need help badly on my homework, but I can do it now thanks to you!

[evacool5]

best tutorial video!!!

[p0izniv111]

Thank you!!!!!! All of your videos have helped me soooo much. <3

[akademikpn0i]

don't mind what all these people say.

i love your videos and they've helped me a TON. keep up the good work! f*ck all the haters!

[WastedGunner]

what is this crap, you are confusing young learners of calculus.

you can easily show that tanx/x -> 1 as well.

you can make this work much easier

[patrickJMT]

ok m'lord.

whatever you say.

[entranceive]

@patrickJMT

Ur vids are very helpful, not matter what these haters say :D

[TheNatalita23]

@patrickJMT haha. nice response.  i think this was most helpful. thanks!

[TheNatalita23]

@patrickJMT

[linkinms]

@WastedGunner I know its been 10months now but sorry you mean you are crap.

[ParkourIvan]

a lot easier solution:

Derivative of sin(x) is cos(x).

at the point 0 you have derivative (growth rate) of 1.

1y/x.

since sin(x)=y

sin(x)/x=1

[patrickJMT]

yes - just use l'hospitals rule.

however you discuss limits before you discuss derivatives.... so... you can not really talk about L'H at this point in the course yet....

that is one big reason for this.

[michalchik]

@patrickJMT You could also do the Taylor Expansion of sin and divide it by X and then take the limit ;-)

[fateater77777]

Oi! give 'em a brake. Limits are taught before derivatives, so for those of us (like me) who haven't gotten there yet, this is the only way we can solve this. So suck it.

[seemagics]

i dont get how tan(6)/1 can equal to

sin(6t)/cos(6t)

can somebody explain this to me????

[IB2009screwed]

tan(x) is always equal to sin(x)/cos(x)

which implies that, tan(6t) = sin (6t) / cos(6t)

[Yu2Kal]

Because Tan(x) = Sin(x)/Cos(x), basic trig identities...

[WastedGunner]

the definition of tan is sin/cos.

or if you think about it from a high school point of view:

opp/adj=(opp/hyp)/(adj/hyp)

aka, tan=sin/cos

[MetallicaFTW7]

Is your Calculus book "Calculus 6th Edition: Early Transcendentals" ?

[clerlic]

About l'Hop and lim sin(x)/x. I remember my lecturer telling something like "you cannot really use l'Hop on that because that limit is used in calculating sin (x)'s derivative". It would be sort of invalid circular proof.

[kasper9016]

so as x->0

sin x ->x

cos x -> 1, am i getting it right?

[mingalabaa]

kasper9016, Sin x is not x. but Sin 0 is 0; Thinsk, for example, Sin 40. Is it 40? No. So be careful. And Cos 0 =1 while as

Cos 90=0; Sin 90= 1

[kasper9016]

oops, thanks for the correction