Limit of sin(x)/x

Proof: sin(x)/x!
Made with Manim. See documentation here:
github.com/Man...
Comment your video ideas down below!

[@trymhc]

But sin x derivative is cos (0) which is = 1, and derivative of x is = 1, (using l’hospital’s rule) ????

[@theoreticalmindset]

Yes your are right. However, this limit is used when proving the derivative of sin(x) so you can't know that d/dx(sin(x)) = cos(x) without knowing this limit.

[@tcmxiyw]

You can use L’Hopital’s rule IF you’ve proven d(sin x)/dx = cos x without using the lim_{x->0}(sin x)/x=1. Otherwise use of L’Hopital’s rule is circular reasoning. This limit is generally used in the derivation of the derivative formulas for both sine and cosine.

[@siraj522]

@@tcmxiyw you can prove that using complex definition of trig functions

[@LeoDaLionEdits]

Squeeze theorem at its finest

[@nomzz1]

as x -> 0 we can use small angle approximations (sinx ≈ x) so the limit becomes x/x = 1

[@theswordslay3542]

a classic proof, though i think it needs a little extra touch to complete it. since we're looking at the first quadrant, we only confirms that sinx/x approach 1 from the right side, in other word this only proof that the right hand limit for sinx/x as it approach 0 is 1. you still had to proof that the limit is the same from both side, you could use epsilon delta definition for it.... OR you could use the fact that sinx is an odd function, which mean sin(-x) = -sin(x), and if we set both x in the sinx/x function we would get sin(-x)/-x = -sin(x)/-x = sin(x)/x. so we got that sin(-x)/-x = sin(x)/x which concludes that if the right hand limit exist as it approach 0, the same limit from the left side would be of the same value. i felt like pointing that out.

[@nomzz1]

if we were to use maclaurin series (sinx = x - x^3/3! + x^5/5! -x^7/7!...) we can see that sinx/x becomes 1 - x^2/3! + x^4/5! - x^6/7!... which as x -> 0 just becomes 1

[@JVsMusicalSoundscapes]

This makes sense since sin x ~ x for x

[@awwab1094]

Very elegant

[@antoniomora1621]

why are there arbitrary markings on the x and y axes not on the circle if this is meant to be a unit circle?

[@theoreticalmindset]

Sorry about that, I'm still new to the program :)

[@TechToppers]

Bro, you used $sin(x)$ instead of $\sin(x) $... Do you think God will forgive you? Awesome video otherwise... 😄

[@mohamedibrahim1023]

Nice vid, but why didn’t you flip inequalities when you took the reciprocal ?

[@theoreticalmindset]

Schoolboy error, sorry about that!

[@mohamedibrahim1023]

Thanks for the reply ♥️