Limit of x*sin(1/x) as x approaches 0 | Calculus 1 Exercises

[@WrathofMath]

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[@Vera-lk1vc]

My native language isn't even English, but I think this explanation is better than the one that could be done in my own language.

[@WrathofMath]

So glad it could help!

[@enochgau1115]

so true

[@akhu123ful]

same here lol

[@punditgi]

Very nifty solution to the problem! The Wrath of Math always manages to squeeze the most out of each lesson! 😁

[@WrathofMath]

Thanks Ezra! It's always my goal to sandwich the best math content I can between the intro and outro!

[@punditgi]

@@WrathofMath 😁

[@Ski4974]

It's actually so helpful how you consider the other possible approaches and explain why it won't work.

[@davidshi451]

I'm finally taking real analysis! You always have a knack of posting videos *just* when I need them :D

[@abdallahabughazaleh9263]

Thank you so much prof. the way you explained every single step and took your time with it was absolutely perfect.

[@WrathofMath]

So glad to help - thanks for watching and let me know if you ever have any requests!

[@applepie4585]

Thank you so much for the clear explanation!! I'm struggling with calculus 1 and this was much needed.

[@MarjoAla-rami]

I love the way you lecture, you explain each step so clearly. Thank you!!!!! So helpful in my analysis class

[@МихаилВеличковский-ы5п]

Thank you very much for doing these videos!!! I think you help a lot of students all over the world by doing this!!!! I'm preparing for my test in maths right now and I'm very glad I've found your channel! Keep up the good work!! 👍👍💯

[@WrathofMath]

Thank you! Good luck on the test!

[@melontusk7358]

This vid just saved my life, dude, you've earned another subscriber. Of course, you deserve as many views and appreciators as 3blue1brown and Flammable Maths

[@matthias2447]

Very well explained, especially the edge cases and your thoughts.

[@WrathofMath]

Thank you! Lots more in my calc 1 exercises playlist; kzbin.info/aero/PLztBpqftvzxUEqGGgvL3EuIQUNcAdmVhx

[@upendrakumarpanigrahi2666]

The way you teach is as awasome as you.

[@WrathofMath]

Thank you, glad to help!

[@uzairwajeeh1737]

By squeeze theorem: lim(x approaches 0) sin(x)/x is equal to one We can express xsin(1/x) as sin(1/x) / (1/x) and taking limit, it should be equal to 1 according to sandwich theorem so a bit confused as to how the evaluation of limit of this function gives zero

[@WrathofMath]

Thanks for watching and good question. sinx/x is 1 as the input approaches 0, but in sin(1/x) / (1/x), x going to 0 makes the input (1/x in this case) approach infinity, so it is a different limit.

[@shachisharma8908]

Hey!...My name is Shachi. Thanks for Solution of each Maths problem!!!

[@statusworld.c]

Hi sachi

[@carolinelotte2156]

just studying for my amendatory test and this helped me a lot. going to watch other of your videos today

[@WrathofMath]

Good luck!

[@WrathofMath]

Studying for the AP Calc exam? Go through a full AP Calc AB Exam with me in this video! kzbin.info/www/bejne/fpvCkpR8mZ6krcU

[@enoq98]

what stops me from doing t = 1/x, x=1/t making me get (1/t)*sin(t) as t approaches 0 making me get the standard limit of 1

[@fatimazahrazine6905]

such a talented teacher ;))

[@WrathofMath]

Thank you!

[@rajpootvikram654]

My teacher had thought a wrong way so I am very sad in this time because he can't understand our future days he pressed for coaching. It is totally wrong for every child. The biggest problem in our India, is coaching. So I pray to god our government must a law for coaching centre.😢😢😢

[@MrAllenmath]

Fantastic explanation! Kudos!

[@WrathofMath]

Thank you kindly!

[@ronakruhal2493]

Nice explanation 👍👍👍👍...but can't we solve the limit such that sin(1/x) dividing and multiply by (1/x) thus it will be equal 1 and x multiply 1/x will also equal to one and the limit of any constant is 1..... So the answer should be 1 ??? If we solve in auch a manner ...

[@fanfsty.]

Hats off to you respect from India;)

[@WrathofMath]

Thanks for watching!

[@faintz10]

your voice permanently altered my brain chemistry

[@WrathofMath]

In a good way?

[@kevinabraham6908]

Just found your channel today great content :) much lov ☺

[@WrathofMath]

Thanks a lot, Kevin! Let me know if you ever have any questions!

[@logeshvar5414]

Love from India ❣️thank you for the Clear explanation

[@WrathofMath]

My pleasure, thanks for watching! If you're looking for more Calculus 1 Exercises, check out my playlist! kzbin.info/aero/PLztBpqftvzxUEqGGgvL3EuIQUNcAdmVhx It's an unlisted playlist right now because it doesn't have too much in it, but I am working on it!

[@logeshvar5414]

@@WrathofMath once again thank ..i am searching for calculus problems only..😅

[@Education-35]

Sometime ask in between the quesⁿs of JEE advanced JEE aspirant 2025 🔥🔥🔥

[@adityabaldha308]

All the best👍🏻

[@Education-35]

@@adityabaldha308 🔜🔜🔜

[@orang1921]

man, that's crazy. love calculus.

[@WrathofMath]

Me too!

[@aanchalsirohi8971]

Thanks from India 🎉😊

[@WrathofMath]

Most welcome 😊

[@anshikaarshi8379]

Thanks a lot. I have a question that can't we write the x which is beside sin(1/x) as : {sin(1/x)}/(1/x)=1 And this way solve? 😅

[@johnsu9949]

nope, you can't use the property lim x-->0 sinx/x =1 here because the angles inside approach infinity and not 0.

[@drofeng]

you can write lim(x->0) sin(1/x)/(1/x) Let u = 1/x As x->0, u->infinity, so we get lim(u->infinity) sin(u)/u sin(u) keeps oscillating while u keeps getting larger, so the limit evaluates to zero

[@Izmoa]

Thanks for the help

[@mergup17]

Thank you so much

[@jiajia1816]

thanks for good concept

[@vaishnavi2612]

Nice explanation🎉❤

[@225hauzel9]

U clear my problem,new subscriber 🌚, keep it up 💪

[@WrathofMath]

Thanks, glad it helped!

[@3467AAA]

Loved it. Thank you :)

[@WrathofMath]

My pleasure, thanks for watching Edgar!

[@zeynepthemartian6423]

how to prove this definition with using formal definition of a limit ? ( it was quite helpful video by the way, subscribed very quick!! thank you)

[@isharauditha4257]

super cool explanation I love it

[@WrathofMath]

Thanks a lot for watching!

[@chanuna]

thank you so muvh for your help!!

[@WrathofMath]

My pleasure!

[@glaratalib4]

lim_(x→0) (cosx) ( sin ( 1/x )) What’s the answer ? I mean limit does not exist

[@nassouroudinenjoya7919]

Merci beaucoup

[@safaa14]

talented 🤗

[@WrathofMath]

Thanks for watching!

[@zeus_rovn]

Thank you so much, and now i understand this exercise

[@WrathofMath]

Glad to hear it, thanks for watching!

[@doolydoko8537]

From morocco ty so much

[@WrathofMath]

My pleasure, thanks for watching! Check out my playlist for more calc 1 exercises and let me know if you have any requests! kzbin.info/aero/PLztBpqftvzxUEqGGgvL3EuIQUNcAdmVhx

[@hhoopplaa]

this was very helpful. one minor question about the method though: if I used squeeze theorem on the absolute value of a function f at a point say c, and by squeeze theorem I found g(x)c) h(x) = L then can I say lim(x->c) f(x) =+-L , i.e. plus or minus L?

[@WrathofMath]

Thanks for watching and good question! No, because it could be that f(x) approaches L from the left, say, but -L from the right.

[@hhoopplaa]

@@WrathofMath Ah makes sense I think. Thanks a tom for the quick answer, didn't expect that^^

[@zeeshannaeem8432]

Tnk you sir.

[@WrathofMath]

Glad to help!

[@takumifujiwara2073]

THANK YOUUU

[@WrathofMath]

Glad to help!

[@odichamathew1700]

Very useful thanx😇🤗

[@WrathofMath]

My pleasure, glad it helped and thanks for watching! Here is a link to my Calculus I playlist if you're looking for more: kzbin.info/aero/PLztBpqftvzxWVDpl8oaz_Co6CW50KtGJy It's currently unlisted because I haven't put many videos in it, but I am continuing to work on it, let me know if you ever have any questions!

[@ayansuleymanova03]

thank you!!! keep going :)

[@WrathofMath]

You're very welcome, I will! Let me know if you ever have any video requests!

[@radhas_reverie]

Thanksss❤

[@sasindudilshan6679]

Very good explanation sir💪

[@WrathofMath]

Thank you! I've got plenty more Calculus 1 exercises in my playlist, and am happy to make more if you have any requests!

[@sasindudilshan6679]

@@WrathofMath please make videos sir ❤❤ they are really helpful 😊😊

[@mohammadhasibemran5955]

did youtube forget to compress ur videos. why is it so HD lol. also ebic video. liekd and subskrieb'd

[@WrathofMath]

Thank you! I record in 4K with a very nice camera - I take pride in going overboard for quality for simple math lessons. "simple" in some ways and not others I suppose.

[@kwpogromca]

I'm wondering if is it scientificly correct to do it a bit diffrent way. I mean of course you cant just substitute x's with 0 because that would lead to 1/0 but if you do it like that: sin is always [-1,1] that implicates the solution is 0 times value from range [-1,1] and that is always 0 So a bit more theoretical approach instead of just stright up calculating.

[@WrathofMath]

Thanks for watching and the question. I'd say that isn't a different way, it's just a less rigorous description of the way we did it. The sine function is between -1 and 1, and the factor of x will make it approach 0. There are different ways we could take care of the squeeze, but the ideas are all pretty similar.

[@joannaaabui]

NICE VIDEO

[@thejahafamily]

Love from Pakistan ❤️🇵🇰

[@WrathofMath]

Much love back from the USA's east coast! Thanks a lot for watching and let me know if you have any questions!

[@thejahafamily]

@@WrathofMath Sure ❤️ with Lots of Love ☺️

[@VioletMemories]

wait, why are we using absolute values? Why is -1

[@WrathofMath]

Thanks for watching and for the question, Miranda! You could have just split this up into a left and right hand limit, show they're both 0, and so the limit is 0 if you preferred. The benefit of the absolute value is that it allows us to avoid splitting the limit into two parts, because whether x is approaching from the left or the right, the absolute value of everything stays the same. So we end up with a proof about the limit of the absolute value of our function. Normally, the absolute value of a function having a particular limit tells you nothing about the limit of the original function EXCEPT in the case where the limit is 0. In this case, we found a limit of 0, which does apply to the original function as well. Again, if the limit of |f(x)| at some point is 0, then the limit of f(x) at that point is 0 as well. I'll do a video on this result eventually, but this is a video I did proving the result for sequences, which may be overly complicated if you're just studying Calc I: kzbin.info/www/bejne/d6PMdp5phqlksJI Does that help?

[@VioletMemories]

@@WrathofMath a bit, but why do we cut out the negatives? Is it because of symmetry and we only need to focus on one side?

[@Browlet540]

Symmetry, ​@@VioletMemories

[@ParaPsychic]

Can you do Cayley's Theorem? There are n^n-2 labelled trees with n vertices, n>=2

[@monky2514]

could you not have multiplied x to -1

[@WrathofMath]

Thanks for watching and good question! The problem with that is that x is approaching 0 from both sides. So, when you multiply everything by x, you are multiplying by something that can be negative. When x is negative, xsin(1/x) is less than or equal to -x. Absolute value helps us smooth over that mess!

[@malakfalkoun1280]

can we use the same method with tan ?

[@taiduan250]

Does this method counted as analyticall?

[@mazlumdemir227]

sorry ı did not get one part of that. What if you multiply the function with (1/x)/(1/x). And ıf you go on you will get x*(1/x) * sin(1/x)/(1/x) and ı got the answer 1

[@MathCuriousity]

oscillates between what and what?

[@marian_ebenbichler]

is this method also possible with the cosinus?

[@WrathofMath]

Yup! Nothing special about sine in this example!

[@tanaka3953]

The answer is actually 1

[@raiza_sabila]

How if lim x->0 x sin (1/x²) sir? Thank you for your videos its very help✨

[@WrathofMath]

Same strategy will work!

[@raiza_sabila]

@@WrathofMath thank you sir🙏

[@raiza_sabila]

@@WrathofMath the answer is same sir?

[@diwakar500]

@@raiza_sabila yes the answer is zero for any positive power of x multiplied by sin(1/x^k)

[@raiza_sabila]

@@diwakar500 thank you sir🙏

[@buiquangtuan1834]

thank you so much 0

[@ahmedelkhomsy2711]

Very good 👍👍

[@WrathofMath]

Thank you!

[@ahmedelkhomsy2711]

@@WrathofMath i have a question lim [x]sin(1/x) x->0 = 0 ??

[@WrathofMath]

I'm not sure what you mean, that's precisely the limit we evaluate in this video!

[@stkhan1945]

@7:28.. .it's getting arbitrarily close to zero ....should it not be ..its getting incrementally closer and closer to zero ? ..instead ..

[@SangTran-ch8wo]

how about limit of xtan(1/x) as x goes to 0

[@WrathofMath]

That one doesn't exist since tan(1/x) = sin(1/x) / cos(1/x). So as x goes to 0, the 1/x terms will approach infinity and will cause rapid oscillation near 0 as well as being undefined at all reciprocals of odd multiples of pi/1, since those values will make cos(1/x) = 0.

[@kumarrajiblovely4490]

Very nice solutions but I am not catch

[@WrathofMath]

Thanks for watching! I'm not sure what you mean, did you have a question I could answer?

[@akkamal6999]

what's the outro music?

[@Nidhisahu123]

I am watching this video in India ❤

[@WrathofMath]

Thanks for watching!

[@nishantmudgal4434]

Why sin (1/x) didn't approached to ∞ at Lim x ----> 0 😑😑

[@WrathofMath]

Thanks for watching and for the question! Why would sin(1/x) approach infinity? Remember the sine function is always between -1 and 1, so there is no way it can approach infinity. The input, 1/x, is going to infinity as x goes to 0, but that doesn't make any difference as far as sine is concerned, it just oscillates between -1 and 1 as always. Hope that helps!

[@nishantmudgal4434]

@@WrathofMath I'm sorry Bruh actually I'm dumb at maths, and was stuck in this doubt yesterday, but i realised what I'm asking doesn't make sence, as i thought The Whole function became ∞, I Didn't Payed attention that only the angle has become ∞, and sine function still lies between -1 and 1. Sorry 😅😅

[@englishbaba6566]

My master said it a wrong method??

[@WrathofMath]

How do they expect you to do it? There are multiple ways, but this way is certainly fine.

[@Zozo-xh7jj]

Can you please do the proof of lim sin1/x x--->0 Is failed to exist

[@WrathofMath]

Thanks for watching! I could do a video explaining why that limit doesn't exist, which would fit nicely in the Calculus playlist. If you want an actual proof, I could do that too in my Real Analysis playlist - that would use a theorem regarding the limits of functions. Something like...look at this sequence a_n that converges to 0, and this other sequence b_n that converges to 0, but notice f(a_n) converges to L1 whereas f(b_n) converges to L2, and thus f(x) has no limit as x approaches 0. So are you looking for an explanation or a proof like I described?

[@Zozo-xh7jj]

@@WrathofMath it was so clear thanks i’m looking for a proof using the definition of limit something like If |x_a| |f(x) -l|

[@reno3955]

Heyyy there Thank you so much for the video I love how you explain all the details ♥️♥️♥️ + Can you please help me with this proof: Lim (x/(2-sin²x)) x->0 I need to write an epsilon delta proof .. I’d really appreciate your help ♥️

[@WrathofMath]

Thanks for watching! That should be pretty straightforward if you work with it a bit. Have delta equal to epsilon, then you'd have |x - 0| < delta < epsilon. But 2-sin^2(x) >=1 and so |x-0| = |x| < epsilon

[@redred5195]

guau thanks

[@neetutripathi943]

Where do you live???

[@shriyasachdeva1024]

the minecraft lucky block-

[@WrathofMath]

Is that what kids call the Mario question block these days?

[@shriyasachdeva1024]

@@WrathofMath maybe

[@shriyasachdeva1024]

@@WrathofMath anyways thanks for the explanation sir

[@alexchan4226]

0

[@williamcorcios1662]

New game: shot every time he says "0"

[@nehayadav8566]

you speak too slow making videos long