limit of 1/(1-cos(ln(sin(tan^-1(e^(x^2)))))) as x goes to infinity
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- Опубліковано 12 вер 2023
- As a calculus teacher, I love creating my own calculus problems and this crazy limit problem is my creation today. I wanted to put this on my Calculus 1 quiz this week but I decided not to because it's only week 3. So I am making a video on it! We have the limit of 1/(1-cos(ln(sin(tan^-1(e^(x^2)))))) as x goes to infinity. What's the answer?
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Thank you all!
You could get the sign just by looking at the outer cosine function. It has to be 1- because cosine can't be 1+ in the real world.
i think he did all the steps for the viewers understand the general reasoning. but youre still right
Ya, that's a lot faster 👍
You are totally correct!
Okay, it's been since 1984 ish, since I did this stuff, but I just put in a few large numbers in my calculator and saw it was going to be +oo . Is that not correct also? Will it bite me?
thats a very good thought man, great work!
I think from just the cos(…) you can say that’s 1^- since cos(x) is always bounded by 1 from above
I was going to say this exact thing. If only I'd seen the video when it was first released :)
@@jimschneider799 beat ya to it!😁
If sin(...) is negative, then ln(sin(...)) would be complex-valued, in which case cos(ln(sin(...))) could be approaching 1 from above.
@@yurenchugood point!
Beautifully explained Sir. plz make more of these simple questions that get people to understand concepts.
Calc 1 student here! I managed to follow along, but the whole 1+/1- thing is new to me. Maybe I missed it, but it feels like a useful detail to keep in mind if it comes up down the line. And now I know how to evaluate whether a limit goes to positive or negative infinity without using a blasted graph.
Kinda cool that I managed to keep up when I would have been deeply confused a month ago.
Awesome problem, much appreciated! After having been shocked for 2 seconds, I pretty quickly came up with 1/(1-1). But I forgot to check for the sign.
I love these kinds of videos with a short fun problem, followed by an explanation!
This one is deceptively easy, and you can also consider that cosine can only ever approach 1 *from the left,* because 1 is its *maximum.*
In general, sin(...) may be negative, which means ln(sin(...)) may be complex-valued, which means cos(ln(...)) may approach 1 from an unexpected direction (or even not approach 1 at all).
this is the type of question that should be extra credit it your grading allows for it
This video helped alot thanks!
Love it! Thank you !!
Nice.i think you are right,this is not for calc test,this is for fun math😀
This is really fun, the real thing about limits!
And you're funny..!
I am so glad I found this channel and I have nothing but pity for that ridiculous algorithm of YT that couldn't show me your content before
Hi Dr.!
This was nifty! You should definitely put it on your test at some point -- math favors the bold.
Calc 1 student here! This isn't very hard, but yesterday my professor asked for a show of hands of how many of us would be taking Calc 2, and out of a class of 24 (community college), maybe 8 of us raised our hands. When she asked about Calc 3, maybe 5. Don't ask for boldness in Calc 1. Many of the students are in it because they tested into it and needed a math elective.
Teacher: The test will be easy
The test: numerator replaced with e^[-(x)^2]
一眼看出无穷😂看分母本以为是一道复杂的Taylor展开题目，但看到分子是1，不存在次数抵消，就意识到比较简单
If we multiply 1+cos(…) on the numerator and the denominator, we have (1+cos(…))/sin^2(…), which is clearly +infinity.
Clever!
Unless sin(atan(e^(x^2))) happens to be negative, of course...
To solve this you have to know how those functions look. That's the real test here.
Pretty cool!
if -pi/2
It doesnt matter which side the ln approaches 0 from, because in both cases, cos is less than 1 near 0.
Why
Even if ln(...) approaches 0 from along the imaginary axis in the complex plane?
@@yurenchu no, if we are in the complex plane, I believe that this limit does not exist because e^z^2 does not have a limit as z -> infinity
@@minifishy7162 cos(x) ≤ 1 for all real values of x, so it could only ever approach 1 from the bottom. Look at a graph of cos(x)
@@yurenchu although it the argument of the cosine was indeed approaching from around the complex plane, you still get 1 but it can indeed be on the positive side.
Pretty cool
a blue pen? that's cheating! 😂😂😂
Nice.
now prove it using M-N def.
That is less than 6minutes and explaining it better than most teachers in 90min 😅
Average Calc 1 Limit
solved in my head before clicking the video 😅
Everything approach from under :) maybe more challenging if u change it up and some approach from above
Y'all are at week 3 while my semester hasn't even started yet 😂
Plz do a M-N proof for the limit now .
I would prefer to change the order of the subtraction in the denominator, thus answer is negative infinity. Reason: If a student answers infinity, how do you know a sign analysis was made, not just doing a faulty solution? Giving the negative infinity answer shows the student did some reasoning to get the correct sign. OR WAS VERY LUCKY.
When I saw the thumbnail i thought it was going to be L’Hopital’s rule, and differentiating that denominator would be a nightmare.
Numerator isn't even 0. No L'hopital's rule there.
@@danielyuan9862 I am very well aware of that, thank you.
1/(1-4pi)
Does the limit exist? My guess is that it goes to positive infinity, and I guess it can’t go to negative infinity since the x^2 means the entire thing is always approaching from the same side
now multiply the denominator by x
The cat t-shirt: amzn.to/3raTywG
A much cooler version is if the numerator is not 1, but e^(-4x^2). the answer is actually finite in that case !
Instructions unclear, initiated L'Hopitals.
should've put 1/x on the top and differentiatied everything
I'm wondering why you chose to use the "minus" "plus" notation and language; I was always taught to use "approaching from left/right" using arrow notation. Fun problem nevertheless :)
It’s just shorthand. I don’t think he invented the notation but it’s also not formal. I wouldn’t tell students with other teachers to use this notation (or do things like put infinity inside the expression) without prior teacher approval.
"Approaching form the left/right' is potentially ambiguous and confusing. For example, what does it mean when a decreasing function is approaching 0 from the left?
Infinity
i finally got a anser wright i just dindt know about the tan-1 thing but otherwise i would of gotten it without help
I think that you should put a similar question on a quiz, because there is a shortcut!
Just look at cosine function (similar for sine), there is no way in real world that sine or cosine can be bigger than 1, so the limit can only approach 1-, not 1+.
A few more people had commented about that.
They got a shorter problem with only two functions 😆
Look again. sin(...) could be negative, in which case ln(sin(...)) would be complex-valued, which would mean that cos(ln(sin(...))) might be approaching 1+ .
@@yurenchu cosh(x) = cos(ix), cos(x) = cosh(ix), min(cosh(x)) = cosh(0) = 1
In order to have cos(ln(sin(z))) approach 1+, we need ln(sin(z)) approaches 0i (almost 0 but on imaginary line), or sin(z) approaches e^0i = 1 ± 0i
Which means z itself has to be a non-real that approaches to real number in order to make that happen.
Correct me, if you can have a valid example.
@@mokouf3 Okay, to put it another way: if sin(z) doesn't approach 1, but for example rather approaches -1 (because for whatever reason z approaches -π/2 rather than π/2 ), then ln(sin(z)) doesn't approach 0 , but rather approaches a complex-valued number (0⁻ + i*(1+2k)π) (with integer k referring to the branch of the ln(...) function). In that case, cos(ln(...)) doesn't approach 1 from below, but rather approaches cosh((1+2k)π) , which (for each valid value of k) is a positive real number greater than 1 ; and hence the limit of the complete expression 1/(1 - cos(...)) _doesn't_ approach infinity.
So perhaps you're right that cos(...) cannot approach 1⁺ (even though we wouldn't really know untill we've actually checked that z in sin(z) isn't non-real). But the point is: cos(...) may _not_ approach 1⁻ , so it's wrong to take the shortcut. We still have to check each nested function in the expression, in order to be able to correctly determine the limit of the whole expression.
why is it not 0 + ?
... the seven deadly sins on your shirt 🙂
The very end of the video is more or less a mic drop. 😆 Or at least, an offscreen mic return. But what better way to conclude a calc solution video.
Please do a rigorous proof on this one now
This would only be a short-answer question on my quiz, no work is required 😆
As a student going to calculus, I’m scared
I’m a bit confused. Once you obtain 1/0, isn’t the limit determined to not exist?
no, because it is a limit not just a fraction, the limit as x goes to 0 from above of 1/x is positive infinity not undefined, and the limit as x goes to 0 from below of 1/x is negative infinity also not undefined
2:30 to 4:50 is useless here: cosine can't approach 1 from above anyway (especially, why do you even draw the trig circle here? what does it bring? from either side of pi/2 for sine or from either side of 0 for cosine, it's the same result)
cos(...) _can_ approach 1 from above, if its argument is complex-valued.
He's explaining Maths in its purest form. Not necessary for problem solving, but great for insight.
All those checks are ok, but unnecessary. Recall: the cosine value is at most 1, thus it follows the sign is plus. 👋🏼
In general, sin(...) can be negative, in which case ln(sin(...)) may be complex-valued, in which case cos(ln(...)) may be approaching 1 from an unexpected direction (or even not approach 1 at all).
Now integrate this 💀💀💀
Couldn’t we just say that cos is always smaller or equal to 1 so the only possibility is +infinity?
no. Only if the value of the cos approaches 1 will the answer be +infinity
Not necessarily, if the argument of cos(...) is (for example) complex-valued while approaching zero.
いきなり式が盛りまくってあるからクスッとした笑
A bit too easy.... I thought we ended up with an indeterminate limit, but no.
How about changing the 0! with 1/x ?
Well. It’s only week 3 for my calculus and we haven’t covered derivatives yet.
Now prove lt 😂
I’m gonna guess that this is gonna be an Euler’s identity heavy video
Try solving in complex world!
U can Directly tell the answer as Infinity🤨🤓😎
Solve x^x=i
the answer is infinity
Les't put 0 istead of the 1 on the top 😆
That's not how you use a lav mic!
dne
Wich religion you are from?
1
limit does not exist as infinity is not a number
I’m gonna guess that this is gonna be an Euler’s identity heavy video
You are incorrect