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for every value in the domain of x²/x, x has the same value. that is to say, x extends x²/x to R. presenting the lesson here as "these are different; there's a hole" instead of "yes, over R\0" feels weird, when the similarities between functional extensions are almost always the important part. I for one think of ln(x) and Log(x) as "the same function, but Log(x) can take complex arguments", and I would be surprised if you don't.

Just use the dominated convergence theorem, move the limit inside the integral, then see that (sin x)^n approaches 0 for almost all x, hence the integral is 0.

Are we not forgetting a possible constant unknown C?

How did you just flip the sin4x/4x? It was in the denominator before. Did you just put a ^-1 sign on it or did you use a property on limits I dont know about?

Seems like you did a lot of extra work Take log of both sides and use log properties to get Log(3) + log(3)log(x) = log(4) +log(4)log(x) Log(3) - log(4) = log(4)log(x) - log(3)log(x) Log(3) - log(4) = log(x)(log(4) - log(3)) (Log(3) - log(4))/(log(4) - log(3)) = log(x) Log(x) = -1 ; x = 1/10. No need to confuse the problem by using base x or base 3/4

I thought that all people who are bit decent at math know this? What's your take?

Spivack's Calculus book bypasses this issue with how they define sin and cos using areas and inverse function, so they derived the derivatives without using lim x goes to 0 of sin(x)/x

These kind of things are very important in rigorous mathematics, when you perform any mathematical operation on an equation, you must respect the number space that the set of equations you produced satisfies for example x=x is only true when x!=0 At least when x satisfies x^2/x=x this is very important to avoid false proofs

…whats the point of the limit then? even if the function appears continuous and has a limit it still fails at zero. which was already readily apparent from the very beginning. there is no scenario in which dividing by zero would be okay, regardless of the continuity and limit. no idea why you are saying its not that simple when apparently, yes, it IS just that simple. dont divide by 0.

It’s also indeterminate for infinite values and for infinitesimal values.

0/0 ❤❤❤❤

That’s why we equate x to zero whenever we cancel x on both sides to solve equations

Why is 0/0 not = 1? You're dividing it by itself.

If the limit exists, I integrate over holes like a bulldozer.

The answer is really simple. Let's say f(x)=x and g(x)=x²/x, for f(x) to be Equal to g(x), 2 things must be true. Df=Dg and f(x)=f(g). In this case, Df=R and Dg=R* so since Df≠Dg f(x)≠g(x). Really liked the video btw, you gave some Very good explanations👍🏻👍🏻

It felt like you were yapping but you were on to something

I can write direct answer by doing these things in my mind because I m asian

Just notice derivative of sqrt(x+1) is 1/2sqrt(x+1) and you want to cancel the 2 in the denominator so you multiply sqrt(x+1) by 2 and get [2sqrt(x+1) + c]

Why use u-sub all the time even on easy integrals? You notice 1/(x+a) is a derivative/function so the integral of that is ln|x+a|. Simply use reverse chain rule if it's 1/(ax+b) and get ln|ax+b|/a

Pls sir revised logarithms products log3log3

Thank you so much! You explained it really well. :)

the chain rule is one of those things where i think its ironically easier to understand if you look at a more complex situation. namely if you look at the situation on manifolds. a manifold is basically just the generalized version of (smoot) curves or surfaces to an arbitrary dimension. if you have a function from one manifold to another, it basically just maps points from the first onto points on the second. and the derivative of the function in a point is then just a linear map from the tangent space in that point (on the first manifold) to the tangent space in its image point (on the second manifold) in that setting the chain rule simply states that the derivative of the concatenation is the concatenation of the derivatives (so the operation of concatenation of derivative commute) so if you concatenate more than two functions, you simply concatenate more than two derivatives. now in the case of functions from R to R, the tangent space is also just R. and the linear maps of the derivatives can be represented by single numbers. (the corresponding linear map would then be t->f'(x)t). and the concatenation of those linear maps would just be multiplication. so if you concatenate a lot of functons, you just multiply all of their derivatives. maybe that explanation isnt helpful in text form. you probably want a graphic for that.

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cos(x)cos(sin(x))cos(sin(sin(x)))

add and subtract one. --> x - ln|x| + c

Honestly I would recommend completing the square.

Students: I will never use calc or the quadratic formula in real life Number Ninja: nuh uh

also would love to have a boss like this lol

2:30 you can just skip the whole quadratic formula by factorising to (n+4)(n-2) = 0 which gets you n = 2 as n cannot be negative.

0:40 You can't just do that without mentioning the fact that the integration interval may contain 1, which isn't defined but doesn't change the integral because the value of the two functions is the same "almost everywhere", in this case, everywhere but 1 point, which has zero measure.

Shouldn't the answer be 3

Yes, the quotient looks remarkably like the formula used in the definition of the derivative. That fact alone doesn't imply that it's improper to use L'Hopital's Rule. There is no circular reasoning involved. L'Hopital's Rule is presumed to have been proven for a far wider range of functions and input values before one comes to solving this problem. We don't have to reinvent the wheel ever time we want to use L'Hopital's rule for f(x)/x as x approaches zero.

Lol...=n...

It can be frustrating wondering if you'll ever use calc in real life. Can you think of a time where YOU might get to use it?

technically using the complex definition of ln(x) the domain is expanded to include the negative numbers and as far as real numbers the imaginary parts which will come with this definition of ln(x) will cancel out in any definite integral as it will always be pi*i, hence in this context where you are allowing logs of negative numbers shouldn't ln(x) be fine without the modulus?

If you break all the problems, you ultimately end up with the axiom that you began with. Also, have you heard of godel's incompleteness theorem? It is easy to get lost in the labyrinth. Or you could just use Lhopital rule and prove it. What matters is 1. Is it a valid rule, whether we are using it properly or not. 2. Whether it is useful or not for the problem at hand. It is perfectly alright to use Lhopital rule. After all, you can find a lot of situations where you seem to go circular with but we don't have to since we already have the knowledge of it. One way or the other, you can do this with many problems. It doesn't mean that the rule is necessarily not useful, implying that it is necessarily not ok to use it despite all the conditions being satisfied.

I really like your videos including this one and they're very helpful! It can be solved in another way too, take numerator= a(derivative of the denominator) + b a and b are constants. It'll also easy

No worries just define the error function as the solution to this integral 😂

I used this method specifically to prove the 68.2% 95.4% and 99.7% of the population under a gaussian within 1,2 and 3 standard deviations respectively, although a taylor expansion solution isn't as nice as an elementary function since it's hard to do things like solve for a value or find the inverse function for example

Yeah but this is, just as you said, an approximation. Besides you can evaluate the integral in the video from 0 to 1 with the same method as the gaussian integral because both are converging to a finite value over the limits of integration. You can take I as the integral from -1 to 1 of e^x² and since e^x² is an even function, your target integral will be I/2. Then you can use any of the methods used to solve the gaussian just with the limits being -1 to 1 instead of -infinity to infinity

i was just doing this yesterday and i found the answer to be sum(n=0,infinity, (x^(2n+1))/((2n+1)n!) ) although i'm not very sure about it

What method did you use to approximate the integral?

1) is because for x < 0, d/dx(ln(-x)) = -1/-x = 1/x. Obviously d/dx ln(x) = 1/x for x > 0 too. Combining the two gives: d/dx ln|x| = 1/x (for x real and non-zero). Hence integrating 1/x gives ln|x|, this is important especially if the domain of x is not restricted to the positive reals only, but also includes negative numbers

Another really useful trick for solving integrals is definitely kings rule, and by far my favourite is feynmans trick. Kings rule basically consists of the following: the integral from a to b of f(x) dx = the integral from a to b of f(a+b-x) dx And feynamns trick (more commonly found in harder questions): the derivative with respect to u of an integral with respect to x = the integral with respect to x of the partial derivative with respect to u -> in other words, you can interchange the integral and derivative sign if you are deriving with a different variable as to which you are integrating. (This is just notation but remember to change the derivative to a partial derivative when you bring it inside the integral)

Another option would be to define sin(x) and cossine(x) with the complex exponencial forms, right? So sin(x) = (e^xi - e^-xi)/(2i), cos(x) = (e^xi + e^-xi)/2

Did you solve this faster than I did? Share your method BELOW 👇

A common mistake even among mathematicians, is not realising the constant can change beyond discontinuities. For example, the proper antiderivative of 1/x is ln(x) + c1 when x>0, and ln(-x) + c2 when x<0. While assuming that c1 = c2 is usually fine, it’s something that people should still be aware of.

really clear , good video

Define trig functions via the integral of area of circle or perimeter of circle and its okay to use the hospital

Ninjas, a few of you noticed an error at 1:15 as the written equation doesn’t match the one at 0:00. My apologies! I hope you still got value out of this video