Calculus teacher vs L'Hopital's rule students
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- čas přidán 30. 11. 2023
- If you are also a calculus teacher, then you know most students will just use L'Hospital's rule for every limit after they learn the rule. So let's investigate the limit as x goes to infinity of (the integral from 0 to x of sin(t)*e^cos(t))/x and see why we cannot use L'Hôpital's rule. This is a great challenging limit problem for your calculus students! #calculus #blackpenredpen #challenge
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Thank you all!
The original problem was equivalent to finding the average value of the function
Oh! that makes so much sense!
You would also get to the answer faster, because the average value of the entire function would be the same as one section.
Very fast and very simple, thanks for the insight
Daaaaaamn, i just noticed 💀💀💀💀
Real ones never forget definition of a derivative 🗣
Chain rule
First principal 🗿
Haha @@NiTiNbRuH
The funny part is that the final step, with 2 and n canceled, is also doable with l'Hopital, in the end
In that part is only a matter of solving the limit, as we got taught in pre-Calculus
lol
@@camilocastillo7245idk what was going on in your school but for me, limits were a part of calculus, not pre-calc
Let's be honest. No student really wants to use the definition of derivative even before they learn the power rule. They do it because we insist that they use it.
First time I encountered L'Hopital's Rule was in a test where the question said, "without using L'Hopital's Rule". I was like, "uh, cool...wasn't going to use that thing I've never heard of anyway, but now I wanna know what it is"
Not knowing what L'Hôpital's rule is while at the same time being able to calculate limits is a power move.
You have gained my respect.
@@BruttiF27to be fair, you’re introduced to limits in precalculus so it’s not necessarily a power move. I get your point though. This example you should know about it since you know derivatives before integrals.
I always put that instruction on my first quiz, over limits. Sometimes there's a student who's taken Calculus before, so I need to warn them away from it.
@@tobybartels8426When I teach Calc BC to students that already did Calc AB (so they learned all of calculus 1 already but we are going back to the beginning for new topics), I tell them they can use L’Hospital but I’m going to be a very hard grader on justification of the rule and using limit notation properly. So most just choose to find the limits the traditional way.
@@BruttiF27 Idk, l'H is only taught in a few countries anyway. It's a weaker and more complicated version of the Taylor-Yound theorem, which we learn right after highschool in France. I think learning learning l'hôpital's rule is a waste of time frankly.
I got 2/pi * sinh(1) using Fourier series
Can you find the Fourier series to the solution of y" + y = tan(x)?
o this is equivalent nice 👍
I remember michael penn doing a video about this topic and the example he gave is x/(x+sinx) as x->inf. This limit is 1 but using L’H rule will give you DNE.
The full L’H rule is that the limit of f’(x)/g’(x) must exist first before we can conclude that the limit of f(x)/g(x) will be the same.
You mean _misusing_ L'Hôpital's Rule will give you DNE!
L'Hopital's rule does not give infinity for your example.
@@isaacclark9825 I never said it did.
If you try L’H rule you will get 1/(1+cosx) but as x->inf this limit is DNE
This one's easy: you can just say that the range of sin is between -1 and 1, and say that as x -> infinity the sine term is irrelevant and you have x/x = 1
@@davidlamas9236 The more rigorous solution is to use the squeeze theorem but yeah that is the idea.
But the point was that if you try to use LH rule since it is inf/inf, you get 1/(1+cosx) which will give you DNE as x-> inf.
And in general, if you try to use LH rule and the limit you get after the differentiation is DNE, we cannot conclude anything about our original limit.
Solving a maths problem : 😔
Solving a maths problem with e or π : 😩
Solving a calculus problem with e or π : 😭😭
Great explanation 🙏🏻
I have to say that from the first video of yours that I watched it was clear that you are a great teacher, and I was impressed and satisfied with your presentation. I actually stumbled on one of your other channels, which led to another, at which point I decided to search bprp to see if there were more channels. That led me to this channel which appears to be your main, though I intend to search for more just in case, and want to check out the content you've been through on the channels I've found already.
I've always had a love for math. Even as a child, before going to school, I would look at my older cousin's homework and try to solve her math equations, and even taught myself how to read and write. I would take every opportunity I could to absorb knowledge in a wide range of subjects, observe and analyze everything I saw, and to think about how things work and how they connect. I even managed to learn basic trigonometry in 3rd grade by accident while doodling in class, during "free time" of course. I never had access to higher education in school and unfortunately didn't do much on my own after school to learn more, but I've always been interested in learning more.
There were many things I wasn't taught, simple and complex, due to transferring to several schools, often in the middle of a term. A video on your channel was actually the first time I've ever been shown long division and, as I mentioned previously, I never took the time or initiative to learn it on my own. I was already aware of the rest of the things you discussed in the video but just from your explanation of long division it made perfect sense to me and motivated me to look through more of your content because I'm eager to learn more about complex math, terminology, and formulas that I never learned.
I tried to get my school to offer a calculus or trigonometry class but they declined saying there wouldn't be enough students participating. 4th or 5th grade was, unfortunately, the last time I actually learned anything at school, in the subject of math. While it made getting a good grade and high test scores easy, I lost my joy for math at some point.
Unfortunately I never went to college, and can't afford to now, but that's never stopped me from seeking knowledge and learning from others. Again, I'm glad I found your channels and I'm grateful for your time and effort you spent into making these videos. You make things easy to understand and I even enjoyed seeing your lessons regarding the things I already knew. Many people are uninterested, impatient, or bored with math, maybe because they didn't have the right teacher or maybe they simply don't care or realize its value but, for the people who are interested in math for fun or for real situations, this content is invaluable.
Thank you for what you're doing and I apologize, to whoever has made it this far, for the long comment. Keep up the great work, I look forward to seeing more of what you do 🙏🏻
This was really awesome. Thank you for doing this one. 🤩
Excellent approach and method, thank you 😁
what a very very nice limit. please make another vedio about this kind of limits
That's a very nice problem illustrating the point. I'm going to try to remember it.
really a good one bro!!!
That was a crazy problem
wow! this was very sharp!
Nice video. You can just do the average height of the square of area e-1/e. Since it has a length of pi, divide area by length to get height. The pi sectors are all the same, to get the area from 0 to x just do (e-1/e)/pi*x. Dividing by x in the limit, you get (e-1/e)/pi. Boom. Done.
very knowledgeable thanks
Very nice, thanks !
Hello do yo have any playlist of learning calculus of your videos ,
it would be very helpful if you share , and my stage is just a beginner need to know every possible thing .
What a great question! I shall remember this!
You have to remember the definition of the derivative; find d/dx of (x^2)(sin (1/x)) at 0.
Nice job!
Neat. I'm wondering if the answer can be found by a different method. What about setting x = 1/u and finding the limit as u -> 0?
Mind blowing fr
Oh my gash, math looks very interesting, when I could understand or learn this? I just trying to figure it out!
I was wondering a long time ago, what will happen when I limit k approach to infinity, and differentiate x to the power of 2k to the k derivative?
Very nice problem. Makes you think...
I never knew that you wrestled my calculus teacher….😊
Hi I love your videos, I’m in math team in my high school and my calculus teacher doesn’t know how to solve this problem do you know how to? k= Σ of 3n/7^n, n=1 and the top of the sigma is infinity.
Was able to figure most of this out without even knowing what the hell l'hopitals rule is or most of what you used in the solution!
I noticed that integral(f(x))/x is just the definition of 'the mean of f(x) from 0 to x', and with that as well as the knowledge that f(x) was cyclical I figured out that in the end the solution was just going to be whatever the mean of f(x) from 0 to π, or in other words 'the integral from 0 to π of f(x) divided by π'!
Thing is I don't really know how to solve integrals outside of like using the power rule so I did have to use your calculation at 3:25, but that's really all I needed to figure out that the final solution was just (e - e⁻¹)/π :D
Good intuition! Being able to see things like this is more important than calculating the indefinite integral (which you could just put into Wolfram Alpha or something).
Are u a physics student?
Coz the mean of function is kinda used more in physics than maths
@@epikherolol8189 Well, sort of
We're still pretty early on and we haven't taught any of that yet, it was just a neat fact I noticed while trying to find out the average acceleration of something for fun once
Brother what brand markers do you use???
Alternate title, calculus teacher versus physics student
Limit as x leads to 0 of xsin(1÷x)
My brain: nah let's use l'hopital's rule
Thanks
Formula for integral 1/X power n + 1 dx with out complex
The same limit but with ln|sin(t)| also works
it's just another Euler's identity (connecting e, pi and -1)! Awesome, dude!
Too bad there’s no i
@@blackpenredpen: Write the final answer as −2i/π×sin(i). (There's still an e in the original integral.)
This problem is so so cool
It's beautiful✨❤
Please tell us you have a sponsorship deal with Expo dry erase pens! :)
Try to find the erf(sin(pi/2)+e-pi
hey man whats up?
Can you find x in this equalition?
x^(x^2+2x+1)(x^2+2x+1)=(x^2+4x+4)?
i saw that you solve hard math problems so i wanted to give you one.I would love to see that you solve this equaltion in one of your videos.
Sir, I'm watching from Bangladesh 🇧🇩.
Where did you find this problem?
####### wow! Does this have anything to do with how you go from discrete Fourier to continuous? I never did understand how that step was possible. The integrals don't seem to converge in the continuous case!
There another example: lim((x - sin(x))/x) when x->inf = 1
It is [inf/inf] but L'H gives us lim((1 - cos(x))/1) x->inf. It doesn't exist BUT its AVERAGE value is 1 too!
How would you recommend going about learning integration from home? Up until expertise
Instead of the alpha thing, I would replace the red n with n+1.
[edit: rather, it's something between n and n+1. pain in the butt to write whether you use alpha or a flexible n or whatever, but anyone who understands limits will see that it immediately evaporates leaving just the n.]
Sir your website is not working for example the question pdf link on 100 integral1-2 can you fix it plasss
How is the top infinity?
Does this problem only exist with x->infinity?
Teacher: Don’t use L’hopital’s rule or else I’ll send you to Le Hospital.
9:28 wait but n is an integer can you do that?
Solve
∫ e^x^2 . Sin (x) dx
very cool
I needed some time to figure that out but exp(+1) - exp(-1) = 2*sinh(1). Probably complex numbers are involved.
I think it's the def of sinh
That's the definition of sinh(x) = (exp(x) - exp(-x))/2
But thats not enought to show that the limit exists. Isnt the definition of the limit of x->∞ that every sequence n->∞ has to converge. So its not enough by showing that one sequence converges?
This is awesome
sir are you from japan or china?Or from other countries?
Always remember!! One of the prerequisites of using l’hopital’s rule is that the limit of the derivatives of the numerator and denominator both exist!!!
(e² -1)/e(pi)
Did that in my head, while listening to music.
Sooooo we can't we use l hospital rule all the time
Why don't we have the area n times?
you can make the area of the curve bigger and still prove it is zero by saying it is less than the integral of 1 since it oscillates between 0 and 1.
edit: nvm, the ratio stays consistent so the final answer is a constant value other than 0.
Pushing to the limit
Can't you sub the absolute value bars for something equivalent like sqrt((x)^2)?
L'Hopital still won't give you a answer
The absolute value bars aren't the problem here. There problem is that the limit of f'/g' doesn't exist here, and that has nothing to do with the absolute value.
to use LH rule u have to prove that the functions are smooth and continuous, top one obviously isnt
A good instructive example. An improvement would be clearer justification of why derivative of integral does not exist.
The derivative of the integral does exist.
d/dx INT{ t=0 to t=x: | sin(t) e^cos(t) | dt } = | sin(x) e^cos(x) | = | sin(x) | e^cos(x).
All you need is continuity on the closed interval of integration for FTC to apply,
and f(t) = | sin(t) e^cos(t) | is continuous everywhere.
Thus d/dx INT{ t=0 to t=x: f(t) dt = f(x).
It's the limit as x goes to infinity of the derivative of the integral that doesn't exist.
lim x--> infinity of d/dx INT{ t=0 to t=x: | sin(t) e^cos(t) | dt } = lim x--> infinity |sin(x)| e^cos(x) which doesn't exist.
I tried calculating the derivative of integral function, using definition of derivative, at points x = n*pi for n=1,2,3,..
and it does not exist.
A condition of L'Hopital's Rule on f(x)/g(x) is functions f and g have derivative existing in the open interval 0 to infinity (in this case). There are an infinite number of points the derivative does not exist. Hence we cannot use L'Hopital's Rule.
I agree the derivative of the integral exists for x not an integral multiple of pi.
@@coshy2748
No, the derivative of the INTEGRAL exists everywhere, including at x = n pi.
You're thinking of the derivative of the function f(t) = | sin(t) e^cos(t) |, which does not exist at x = n pi.
But the derivative of the function F(x) = INTEGRAL{ t=0 to t=x: f(t) dt } exists everywhere.
In fact, F ' (x) = f(x), so F ' (n pi) = f(n pi) = | sin(n pi) e^cos(n pi) | = 0.
Again, F ' (x) = f(x) is directly from the Fundamental Theorem of Calculus, as f(t) is continuous everywhere.
I eyeballed it immediately lol
How about the integral from 0 to infinity (x^-1)dx?
I tried for so long and couldn’t find a way to solve it, however there definitely is a solution, according to Desmos
Integral of 1/x is ln(x)+C but integrating from 0 to inf won't give you any actual results cuz ln(x) tends to infinity as x grows, and when approaching 0 it goes to -infinity. You might be misinterpreting desmos.
@@Kassakohl So are you saying this integral diverges?
@@danobro yes
Which of the following sequences could represent the impulse response of a stable discrete-time system?
k^2
(-0.65)^k
2^k
ksin(k)
Doyle's constant for the potential energy of a Big Bounce event: 21.892876
Also known as e to the (e + 1/e) power.
At the eth root of e, spaghettification of particles smaller than the black holes. Other than the relatively small amount of kinetic energy of black holes being flattened into dark matter, the only energy is potential energy, then: 1 (squared)/(e to the e power), dark matter singularities have formed and thus create "bubbles", leading to the Big Bang part of the Big Bounce event.
My constant is the chronological ratio of these events. This ratio applies to potential energy over kinetic energy just before a Big Bang event.
that is definetly something i guess???
Sin(inf)=?
I have a question about a thing I saw in an older video of yours. In said video you stated that sqrt(-x)=1 has one solution, -1. This is correct, don't get me wrong, even Wolfram Alpha says so. However, why is it that when you do sqrt(-1)*sqrt(x)=1 and then you replace sqrt(-1) by i you get no solutions? Are we not supposed to do that?
I dont get it. It should still give -1 right, i*sqrt(x)=1, so x =(1/i)² which is -1
@@fahimnabeel606 That is correct... Until you replace the x in the original equation to check: i*sqrt(-1)=1 i² = 1 -1=1, which is impossible. This is giving me a huge headache.
If you’re working with real numbers only,
the rule sqrt(a*b)=sqrt(a)*sqrt(b)
holds only for a and b being positive reals
If you’re working with complex numbers,
You can extend the rule
Sqrt(a*b)=sqrt(a)*sqrt(b)
If:
•a and b are both positive reals
•a and b are reals with opposite signs
However, when a and b are both negative reals, this rule doesn’t hold anymore
In your case, you tried to apply it with a=-1 and b=-1 which is why it leads to a contradiction
It is said that in order to avoid this « mistake », we choose to write sqrt(-1) as i instead (so we don’t think of using properties of radical that don’t necessarily hold in the complex world)
@@goblin5003 yeah, that makes sense. The way I was thinking i couldn't exist because sqrt(-1), when squared results in -1 if you cancel the root with the square (correct way) but if you multiply both -1 in the roots, which you can technically do because of same exponent, you get sqrt(1) instead. Thank you for clearing my mind.
can we prove that the resulting ratio is transcendental?
I think it's super super cool,that's it.
Can you help me solve this question algebraically:
How many real solutions does the equation 2^x+x=0 have? By the way I don't want to use the product log function. Please justıfy each step
To begin, we can rule out all positive values of x as real solutions, because 2^x is always greater than 1 for x>0, and x>0 is trivial. This implies that 2^x + x is always >=1, when x>=0.
Next, what we can do, is show that the derivative is always positive.
d/dx 2^x = ln(2)*2^x
d/dx x = 1
d/dx (2^x + x) = ln(2)*2^x + 1
Observe that this function is continuous and differentiable at all real values of x, so there are no sudden changes in direction, not accounted-for by a derivative.
Exponential functions are always positive, for all real inputs. ln(2) is also a positive number. This adds up to the derivative always being >=1, for all real values of x.
This means that once the function crosses the x-axis, as x is decreasing, the function doesn't turn around to have a negative slope. A lack of a stationary point, and a lack of a negative derivative, means that the function is always decreasing as x decreases. This tells us that there is only one real value of x, where 2^x + x = 0.
Finding the solution does require the product log function, but you can show that there is only one real solution without using product log.
sin(t) e^(cos(t)) is periodic -> the limit equals 1/(2pi) • integral of |sin(t)e^cos(t)|dt from 0 to 2pi. sin(t)e^(cos(t))dt=-d(cos(t))•e^(cos(t))=-e^(cos(t)) -> -e^(cos(pi))+e^(cos(0))-(-e^(cos(2pi))+e^(cos(pi))=2e-2/e. The limit is (2e-2/e)/(2pi)=(e-1/e)/pi.
Is this computer science?
hello a question :
There are two concentric circles with different radii. Lines (rays) drawn at a certain angle from the center intersect both circles at one and only one point. That is, for every point in the small circle intersected by the line, there is a point in the larger circle. but the outer circle is bigger!!!! how is it? Will some points in the outer circle remain empty?
It is correct. There are as many points in the smaller circle as there are in the bigger circle.
Another example is that there are as many real numbers between the interval (0,1) as there are in (1, infinity) because there is a bijection between the two sets ( i.e. f(x) = 1/x )
This amount of things is called the Continuum. It's the cardinal of the real numbers R (and also any interval in R) and since simple geometrical shapes are just bijections from an interval into the plan of course all shapes must have the same cardinal (i.e. amount of things)
I forgot to mention why you might find this though experiment strange is that "amount of things" and "measurement" are different things when you want to make sense of things like these. Measure is an intuitive notions: the lenght of an interval should be the difference between it's endpoints( the lenght of the interval [2,5] is 5-2 = 3) or how the area of a unit square is 1x1 = 1.
Not to confuse yourself thinking that any shape, as long as its got Continuum of points has lenght (perimeter)! A favorite weird example example of mine where measure and cardinality don't seem to match is Cantor's set:
Take the unit interval [0,1]
Remove the middle third of the interval and we are left with [0,1/3] U [2/3, 1]
Repeat this, remove the middle parts of these guys and we get [0,1/9] U [2/9, 1/3] U [2/3, 7/9] U [8/9, 1]
Repeat this process of removing thr middle ground of each smaller interval forever.
Cantor's set is the limiting set you get after all of these.
What's interesting about it? The cardinal(amount of elements in the set) is a Continuum(why? think about it), yet the measure(lenght of the set) is 0 since we keep removing thirds of the set again and again, the limit lenght of the set must be 0.
So a Continuum set yet with lenght 0?!
And there are even more weird such counter-examples to our normal intuition, if you are more interested in this search for "Measure theory", although I warn you it's not for the mathematicaly weak
Please solve this integral: I(x)=e^(ax)tan^(bx). Thank You.
Is it some standard question or u just made it up?
e^(ax)sin^(bx), e^(ax)cos^(bx) integral solution exists. So, why e^(ax)tan^(bx) don't exist??? I have tried many ways to solve this but I can't. But I need the solution of this problem.
@@akshatkachave108
Leibnitz theorem go brrr
Ah yes, e raised to the cost power… but at what cost?
I can't write integration question in comment i want to give an image where should i upload sir? Thanks for teaching us sir ❤.
Look for a service like Imageshack, that lets you host images, and it will generate a link. Then post a link to it in the comment.
You also can use standard in-line notation to provide the math expression. Use the up-carat (^) to indicate exponents, the asterisk (*) for multiply, the slash for division, the underscore (_) to indicate subscripts, and plenty of parentheses and brackets to clarify intent. You can type out the name of Greek letters, or go to Lexilogos and type them by typing Latin counterpart letters.
cleannn
Solve 200 linear equations challenge for you 😅 you must accept this
Please solve/extend (x+y)^z (z is REAL)
Hi brbp, good video! I have a two videos suggestions:
• All solutions of the equation sqrt(x^x) = x^sqrt(x)
• Any easy method to solve the integral 1/sqrt(2x^2 + 1)
Ps: I wanted to see a solution other than the trigonometric sub, there is probably one with complexes or contour.
I think the second one can be done in the usual way, by closing the contour with a halfcircle in the upper plane around the origin, with radius going to infinity.
You can definitely do the second one with complex numbers and partial fractions, although I'm not sure that it's really _easier_ like that.
sqrt(x^x) = sqrt(x*x*x*x*x*…) [x times]
= sqrt(x)*sqrt(x)*sqrt(x)*sqrt(x)*… [x times]
= sqrt(x)^x
sqrt(x^x) = x^sqrt(x)
sqrt(x)^x = x^sqrt(x)
Can’t be bothered right now, I’m tired
[ ∞ = π/2 (8) ]
Or
[ ∞ = 90˚(8) ]
How do we know that x is greater or equal to 2π? What if x is less than that?
X approaches infinity in the limit, so of course it's bigger than 2π. But even if it wasn't, the n of the expression can also be 0, thus the statement remains true.
@@Nostale97 thanks alot
Can someone tell me why we can ignore the alpha(area) and constant k?
Cuz they don't matter!
Jk I don't know either
Yea, I figured it out.
As n is approaching infininy, 2(e-e^-1)n becomes a very large number. As alpha is less than (e-e^-1), is becomes an infinitely small value, as compared to 2(e-e^-1)n, so alpha can be considered negligible.
Same is with the denominator.
As n approaches infinity, k becomes infinitely small as compared to 2πn, so k can be considered negligible as compared to 2πn.
So, both alpha and k do not matter in the limit.
@@lakshya4876 but does it affect the limit in any other way? maybe the answer get wrong even though they are negligible
@@saravanarajeswaran2626 no, this is a common theme in limits. As one value approaches infinity, a small constant becomes negligible.
Consider the following limit:
*Lim (1+n)/n*
*n -> Inf*
We can split this into
*Lim(n->Inf) (n/n)+(1/n)*
= *Lim(n->Inf)* *1+(1/n)*
Now, *as n approaches infinity*
*1/n becomes infinitely small* , nearly equal to zero.
So 1/n can be considered negligible.
So,
*Lim(n->Inf)(n+1)/2=1*
@@lakshya4876 yeah i know this but still, this doesn't apply to that sum does it?
lim (f(x)/g(x)) x->inf = [inf/inf] = lim((f(x)/x)/(g(x)/x)) x->inf = average(f'(inf)) / average(g'(inf))
Is it true? My conclusion make me dizzy a little bit.
Can you do tan(x)=i ?
Undefined. Arctan(i) and Arctan(-i) are branch points for which connect the branch cuts along the imaginary axis until infinity.
What about the k at the end ? Oupsyy
Int [ | sin( t ) e^(cos ( t )| dt]=
= - (1/sin ( t ) | sin ( t ) *e^(cos ( t )|+C
😅
I am from bangladesh and i love math🥰
Holy
It would be pretty cool if solve me the following question which I found and I could not solve.
limit x approaches 0 of (x^x^^^x -x!)/(x!^x! -1)
Bruh I can't even understand what you have written
Maybe better notation would work
Wait what? Isn't the limit supposed to be non-existent?
Calculus teacher always wins