Calculus - How to find trigonometric limits using sin(x)/x
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- Опубликовано: 30 сен 2024
- Trigonometric limits involving sin(x)/x can be very tricky. Check out this video to see how you can manipulate functions in the just the right way so you can use this great tool. For more videos please visit www.mysecretmat...
THANK YOU, WHY THE WAY SOME PROFESSORS ARE TEACHING, MAKE IT SO HARD TO UNDERSTAND. IN LITERATELY 7 MINUTES YOU TAUGHT 4 HOURS WORTH OF TEACHING.
Par tan
Same!
They make it so complicated
This is honestly unbelievable. I don't know why people like you get on RUclips and bless the public with your knowledge and manage to make it SO understandable. Thank you so much for you efforts, and please know that you are helping so many determined students out there!
I love you so much right now!!!! I've been struggling with this and you've made my morning better. Thank you!!!
same
Same
You just explained this concept way more simply, way more clearly, and way faster than the university profs that I pay to teach me. Thank you for ending my struggle with these kinds of limits!!
Midterm in 4 hours, completely forgot this section existed. THANK YOU SO MUCH FOR THE HELP!
My brain doesn't hurt anymore! For that, I thank you!
I can tell by your voice and other aspects of how you talk, that you would do an amazing impression of Jeff Goldblum. You should try it. Assuming that MySecretMathTutor is not actually Jeff Goldblum.
This seems too easy, and understandable....somethings wrong.
these problems are a lot easier than the ones i see on my test but it still helps :)
Thank you for this! I couldn't seem to wrap my head around by multiplying by what you need and how you could take numbers out the front of the limit. You have saved me from embarrassment next math lesson, thanks!
you are a lifesaver....really I like your way how you do every topics...
Oml thanks for this video! Very clear, covered all the possibilities of the things I was worried about
Loved ur way of teaching
you're awesome!!! I was so worried about this chapter, i have an exam tomorrow. but now i have no worries!!
THANK YOU
This is very elementary. The real problems are way too hard to solve using only this deduction.
For instance, what's the limit of (1/x*cosx) as x->0.
vipul jhamnani It's one over zero so it's infinity.
The limit doesn't exist.
Ive been looking for this exact way of how you taught this for a half an hour. You definitely taught me more than my prof did in three hours. so helpful.
plz solve the following question for me
lim x->0 sin(1/x)
thank you so much! Honestly i have learnt more from 3 of your videos than I have going to my lectures. Definitely one of the best!!!
Thanks that was really helpful :)..... just want to add sth. There is a rule for a question like (tan 5x ÷ x)
It is that (tan x ÷ x) = 1 when x→0. In that question u could just make the denominator equal to the neomonater by the multiplication of 5.
This helped a lot even after 9 years, thank you!
Glad I could help out. Keep up the good work. :^D
Thanks so much! Even after my prof's lecture and having watched several people's videos on YT, I was still unable to understand this until now. It seemed silly that I could understand all other limits, derivatives and rules, etc. but could not get this! I felt so stupid.
I was having so much difficulty with this formula but with your explanation I actually understand it now!
SO how is (sin5x/5x=1) I DONT UNDERSTAND!!
It's a property you can use. sinx/x=1 you just have to make sure the numbers correspond, in this case we have 5 on both sides so it still equals 1.
Mikis Mays If you make a graph it actually does approach the value of one.
But you have a denominator as a 5 x so won't it be undefined after you take the limit as x approaches 0?? At 06:31 I see sin5x/5x *1/cos5x so how is that sin 5x/x not a does not exist? SMH......help.
You have an amazing way to save us all from Calculus! 👍👏
great teaching! thank you so much
at 10:40, how can 4x / sin4x be 1? If you plug in the 0 in the denominator shouldn't get 0?
Thanks for this very helpful video. I have a question: You mentioned that it is OK to separate a constant value and multiply by it later. I noticed that when you pulled out the 5 (Numerator) it became a 5 (which is the same as 5/1), then later when you pulled out a cos(5x) (denominator) it became a 1/cos(5x). Is this a general rule, that when separating denominators you must keep them as 1/n? i.e that would make the second example of 11/3 a shortcut for pulling out (11/1) then pulling out (1/3) and then multiplying to get 11/3.
It does very much, thank you for the reply.
Thank you, this was as clear as day
really awesome thanks for providing help.
Thank you so much. You clarify it simpler for me.
thanks bro
ur the real map bro
*mvp bro
do we always multiply by 1 at the end? no matter the equation?
THANKS!!!!!!!!!!!!!!!!!!!!
THANK YOU
You helped me to carry out a math lesson in English.
Thank you so much, this is excellent!
Why do we want the same on the bottom as on top? Unclear
I have hard time understanding why lim(x>>>0) of sin(x) is not 0, because obviously the sinus of 0 is 0, and that's what it is constantly reaching for. This also messes with my brain because obviously then lim(x>>>0) of sin(x)/x would be 0/0, but for some reason its considered 1, and even graphically I dont understand why it is 1
thank you so much for helping me
so when you have the problem lim when x is approaching 0 of tan(5x)/x when you split everything up is it ok to say you're converting the 1/cos5x and saying that 5x = y inputing a 0 into the y and cos of 0 is 1 so 1/1 = 1 ? sorry hope i wasn't too confusing explaining what i think you were doing.
You're not writing it in the right notation. You need to use parenthesis because I got so confused as to why 1/cos5x was =1 and why not 5. instead it was 1/cos(5x)= 1/cos(0)=1.
ughh go away
4:35 how did you move the cos5x down?
Why is sin5x/5x equal to 1?
for the one where the answer is 1/9, wouldn't you multiply the cosine of 0 (which is 1) times 12, giving you a final answer of 1/108?
Why you didn't show limit sinx/x =1 as x goes to zero?
God bless your kind soul
Ok I will be honest straight forward this time
How tF did you get (4x/sin4x) as limit of x approaches 0 to be 1 All I know is 4x0=0 and sin(0)=0 >:'D besides you didn't even do that strange theorem when you divided the leading coefficient in brackets to numerator and denominator >:"D!
help~
So basically if cosine is alone on the bottom it just can be changed to the 1 because it’s limit will always be approaching 1?,
With these, you are trying to focus on pieces that become "indeterminate." Think of things like a number divided by zero, or zero over zero. In these situations we handle them using value for the limit of sin(x)/x. Down the road we can also make better sense of these using L'Hopitals rule. :^D
There are limits that also involve 7 indeterminate forms (0/0, plus or minus infinity/plus or minus infinity, infinity-infinity, 0*plus or minus infinity, infinity^0, 1^plus or minus infinity, and 0^0) using l'Hopital's rule. This usually works if it cannot be done using algebraic method. The example in this video will work since plugging in the value will be 0/0. You can take the derivative for both the top and bottom separately (not the derivative for quotient rule). The result will be limit as x goes to 0 of cos(x)/1, so cos(0)=1.
sir pls explain the problem when x tends to infinity sin x/ x. I tried doing this in the following way. let x = 1/h so when x tends to infinity, h tends to 0. so now lim h tends to 0 sin (1/h)/1/h. according to your video the answer should be 1 but the answer is 0. plz explain.. Thank you..
you lost me at 5:17 LOL.The Organic Chemistry Tutor explained this part pretty well in his "Limits of Trigonometric Functions" video @ 10:09. But as for the rest of the video, you are the simplest out of everyone I've watched. THANK YOU!
Sometimes it's best to hear a topic from multiple sources. Great job on finding some great instructors. :^D
How would I go about solving the following: x approaches 0 sinx/8x In other words, when the number is only in the denominator.
You can think of the 8 as a fraction separate from the rest....
(1/8)*(sinx / x)
When you take the limit as x approaches 0, the 1/8 is a constant and will be unaffected.
So you are left with
(1/8)*(1) = 1/8
Hope that helps out. :^D
God bless u.. maan.!!
This video is great!!!!
The way you explained it all made it very easy.
I can't thank you enough!
Great video, you explained so clearly! I’m curious, though, is there a reason why you can’t just use L’hospital’s rule to solve these?
I'm A slow person I'd like to know how you get to the 11/3. Sorry but this didn't help
what about
limit as x goes to 0, (sinxcosx)÷4x
?
I'm really confused
as limit of cos x is 1 why cos 4x/cos4x is not 4/4 i mean still 1+(4/4)=2 and is is cos4x=1??
christos christou Cos(4x)/cos(4x) is one just so the limit is just one. 4/4 and cos(4x)/cos(4x) both have the same value so both approaches the same value which is one. And yes if the limit approaches zero then the limit of cos(4x) is just one because cos(0) is just 1.
Bless you for your neat and helpful video clips!
I wish to rapidly find the field in which I can repay the community in a level of contribution comparable to yours!
You made my calculus homework 100x easier. I LOVE YOU
SIR I THINK THAT IAM VERY BACKWARD IN MATHS.BUT I COULD FINISH ALL THE PROBLEM CORRECTLY THAT YOU HAVE COVERED. THANK YOU
Thankyou very much!
what about lim as x approaches something other than 0? like pi from the left for example
very helpful. i do hope you give exercise for every lessons but thank you so much.
i tried to understand this for about 3 months and this video could just make me understand under 20 minutes. thanks a lot=))
YOUR THE SHIT BRO THANK YOU
Wow....Thanks a lot. From now on, you are indeed my secret math tutor...
WOW THANK YOU
how about this problem, x^2/sin^2(3x)? i get what you saying
Sir as Gujarati ma rakhay and bolay to vadhare Sanjay under stand all people clearly so please speak Gujarati in your video
very very good thank you
Thank For Teaching.
what if x is approaching not 0, like 1 or infinity??
QUESTION
If you *factor* out the 4x, shouldn't the *numerator* now read
4x(1+cos(1)) ???
Can someone please tell me what rule was used to keep the cos(4x) unchanged after factoring?
4x+cos(4x) = 4x(1+cos(4x)/4x)
You can't move out the (4x) inside the cos or sin.
limx→0 (2*sin^2 x)/(tan 3x*sin 4x)
how do i solve this?
This video was beyond helpful!!! Thank you for taking the time to make it!
Wow! Your way is so simple!!!!!!!!!!!!!! I love the Internet.
I think its easy to suppose limx→0tanx/x = 1
MAy God bless u forever...... I understand these questions now :3
god. thanks a lot!
Excellent video thank u very much .
Thank you sir🤩
This video was really helpful. Thank you!
I appreciate this so much ! Thank you !
bro can i marry yu?
Ok I see you saying thats how its supposed to be. :v
Very helpful! Thanks so much
Bless your soul. Time and life saver
x aproches to pi then sinx/pi-x is equal to
الله اکبر
damn... I had one example on this..
شكرا استاذ
Really Thank you , you help a lot , keep it up ;)
Thank you it is really useful👏👏
Thank you. I have been struggling wiz zis.
🖤 my brother
I love you. Incredibly helpful.