derivative of sin(x) by using the definition of derivative 

For your knowledge, both the limits are famous limits and they do not need the derivative of sin(x) to be proven. You can use the Squeeze Theorem to prove both.

I agree. I was waiting for him to derive the limit as h ->0, which he did not do.

Squeeeeeeze.

genius.

yeah

you do not need to use l'Hopital's. You can prove these without that. But they're a bit complicated

Dirk van Duivendijk A bit late, but you can prove sin(x)/x -> 1 by comparing areas on the trig circle, transforming the expression so you get sin(x)/x in between the two inequalities and then using the squeeze theorem to deduce the limit

Well, you can use Squeeze Theorem and you’ll see on the graphs they approach 0 and 1 respectively

sin( reallllly small x) = 0

Excuse my english. It's circular because you assume to the derivate of sinx is cos x for the second term, then the derivate of cosx is -sinx for the third term, ...

@@我妻由乃-v5q so take the whole derivative on all terms . which leads to cos (x)?

@@raifsam3408 bruh the whole point of the thing was that you can't use the derivative of sin to prove the derivative of sin

@@angelmiranda-acosta6092 that's what i asked the person up there.

@@raifsam3408 the Taylor series uses the derivative, so using something that uses the derivative of sin to find the derivative of sin is wrong

Yes it easy for solve it

That is one shortcut from the limits of trigonometry. A video from Khan Academy explains it well here. www.khanacademy.org/math/ap-calculus-ab/ab-limits-new/ab-1-8/v/1-cosx-over-x-as-x-approaches-0

if you want to know a limit in a natural way, you just have to approach the x from left to right with increasingly shorter values and you will find the limit in the purest way, and you gonna see that the limit is 0 without any LH's method or whatever

Just do the following procedure Lim ((cosh-1)/h) = Lim ((cosh-1)(cosh+1))/h*(cosh+1)= Lim (cos^2(h) - 1)/(h*(cosh+1)) = Lim(sin^2(h))/(h*(cosh+1)) = Lim (sinh/h) * Lim((sinh)/(cosh+1) 1*((sin0)/(cos0+1))= 1*(0/(1+1))= 1*(0/2)= 1*0= 0

He did a very good presentation. I have this problem with these proofs on my channel ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-64dguvQBwUQ.html

why?

@@realDewSplatGum Its also cosinus in Hungarian.

@@realDewSplatGum I wthink it's cosinus in every language but english.

@@qqtrol1774 no, it's not. In German it's "cosinus", shortened to "cos(x)"

That's why we call it ch here instead of cosh

@@my.lionart "It's not cosinus, it's cosinus"

I want to ask how happen that the limit of (cos h - 1)/h as h approaches to 0 becomes 0 without using L'Hospital rule?

It's a good question. If we are calculating the derivative of a function, we can't use the derivative of that function because that would just be circular. We have to calculate the limit without the need for the derivative of cos. (Ok, technically speaking the video is calculating the derivative of sin(x), not that of cos(x), but I think in the spirit of the problem we shouldn't be allowed to "know" the derivative of cos(x) without a proof either...)

agree with u warpRulez

+jonathan enriquez Here is a simple proof (using only the fact that sin(x)/x → 1 as x → 0). Well, -- note that I've used 1 - cos(x) instead of cos(x) - 1 ... since the first is positive -- of course it makes no difference with respect to the limit [ 1 - cos(x) ] / x = -- multiply both numerator and denominator by 1 + cos(x) = [ (1 - cos(x)) * (1 + cos(x)) ] / [ x * (1 + cos(x)) ] = [ 1 - cos^2(x) ] / [ x * (1 + cos(x)) ] = sin^2(x) / [ x * (1 + cos(x)) ] = sin(x) * [ sin(x) / x ] * [ 1 / [ 1 + cos(x) ] ] -- this is already enough to prove the statement -- but see below for an additional statement on 1 - cos(x) -- (hence the extra factoring out of 'x') = x * [ sin(x) / x ] * [ sin(x) / x ] * [ 1 / [ 1 + cos(x) ] ] = x * [ sin(x) / x ]^2 * 1 / [ 1 + cos(x) ] Since sin(x)/x → 1 as x → 0, it is easily seen that the limit indeed approaches 0. Actually, we have proven a bit more here: lim (x → 0) (sin(x)/x) = 1 implies that sin(x) ≈ x for small x. If we plug that into the expression above, we get (1 - cos(x)) / x ≈ x * 1^2 * 1/(1+1) = x / 2, hence that 1 - cos(x) ≈ x^2 / 2 This result immediately follows from the Taylor series expansion of cos(x). (which we should not use at this point, since we do know neither the derivative of sin(x), nor that of cos(x) yet) Now, using the fact that sin'(x) = cos(x), we could either derive that cos'(x) = -sin(x) by using trig identities for cos(x+Δx) (and the limit above), or use the fact that cos(x) = sin (π/2 - x), together with the chain rule.

Matematician showed one way to calculate lim h->0 (cosh-1)/h but there is other one Use double angle formula and if necessary pythagorean identity to write numerator in terms sine of half angle We can also write numerator as cos(x)-cos(0) and use sum to product formula

yeah

yes thats what i was thinking, since the left hand limit and right hand limit both equal zero, the limit as h approaches zero, is zero

I mean technically yeah, but If we know it’s continuous then the limit has to exist

🤓

They both approach the same thing.

Wow Cool Guy, that was very cool! Thank you for sharing!

Nice if all you want to prove is the sine derivative (and you're not picky about 1-sided versus 2-sided derivatives). You miss the fun of proving limit as h -> 0 of versin(h) / h, though. Hint: 1 - cos(h) = 2sin(h/2)^2.

Nope, CORDIC algorithm

+Couperin le Grand The fact that lim (h→0) [(f(x+h)-f(x-h))/(2*h)] exists, does not necessarily imply that lim (h→0) [(f(x+h)-f(x))/h] exists. (although it does hold for the function sin(x) - see below) As a counterexample consider the function g(x) defined as x*sin(1/x) when x≠0 and 0 when x=0. This function is continuous everywhere and differentiable for every x≠0. Also, since g is even, it follows that lim (h→0) [(g(0+h)-g(0-h))/(2*h)] = lim (h→0) [(g(h)-g(-h))/(2*h)] = lim (h→0) [0/(2*h)]= 0. However, lim (h→0) [(g(0+h)-g(0))/h] = lim (h→0) [(h*sin(1/h)-0)/h] = lim (h→0) sin(1/h) and this limit does not exist. On the other hand, IF a function is differentiable for x=a, then it IS true that lim (h→0) [(f(a+h)-f(a-h))/(2*h)] exists and that this limit equals f'(a). If a function f is differentiable for, say, x=a, then the difference quotient [f(a+h)-f(a-h)]/[2*h] gives a very accurate approximation for f'(a) (error of order h^2, if f'''(a) exists), which is the reason why it is heavily used in numerical analysis. Note: the functions g(x)=abs(x) can also be used as a counterexample, but the set of functions of the form g(x)=(x^α)*sin(1/x) when x≠0 and 0 when x=0, where α>0, forms a more general set to (dis)prove certain assumptions with respect to (especially) differentiability.

Thanks.

Have in mind lim as x -> 0 of senx/x is equal to 1. Knowing that you can calculate both límits.

Wikipedia has a nice instructive article on how to do this limit without LH. Basically, you need to consider the unit circle and some areas.

There's also the squeeze theorem for sinh/h

It would be circular reasoning

?

Glad to hear : )))

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I don't think it's an actual proof though. When he took the limit of sin(h) / h as h -> 0, he said it's 1 but the way you solve that limit is using L'Hopital's rule and taking the derivative of the numerator and denominator. But when you take the derivative of the numerator, you are taking the derivative of sin(x), assuming you already know it's cos(x), but that's what this entire proof is trying to prove. So you are proving derivative of sin(x) is cos(x) while already assuming it true in the proof, so it doesn't actually prove anything.

You are missing grouping symbols.

Me japanese. i know, China has some talented math/study/science proffesionals. But, in general, chinese people are NOT good at math/study/science. Because... The fact they have 1.5 billion people and at No.2~3 of GDP in the World. and China still had been achieved very few (of The Fields medal , Novel Prise). in logic, if they are good at math/study, they Must had achieved 10+ these medals.

ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-64dguvQBwUQ.html

you haven't really expressed why (1-1)/h goes to 0 as h->0. That is, you haven't shown that cos(h) goes to 1 faster than h goes to 0. this is a bootstrapped argument

Uchiha Madara here's a slightly better approximation. Cos(h)≈ 1-h² for small h, so it would be lim h→0([1-h²-1]/h) = lim h→0(-h²/h)= lim h→0(h)=0

engineers be like

you damn engineer

That's a circular argument. The reason why sin(h)~h for small h is that lim sin(h)/h=1 as h->0

Doesn't the Taylor series use the derivative though? It.would be circular reasoning

@@angelmiranda-acosta6092IK,i am just writing a good note it doesnt nessecerly answer the question right but i liked to write it

ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-64dguvQBwUQ.html

Jacek Soplica lol how can you use pythagorean identities here

cos(h)=cos^2(h/2)-sin^2(h/2) - double angle formula 1 = cos^2(h/2)+sin^2(h/2) - pythagorean identity cos(h)-1 = -2sin^2(h/2)