sin x / ( 1 + cos x ) = tan ( x / 2 ) so you can also evaluate it as : lim ( x -- > 0 ) sin x / x * lim ( x -- > 0 ) [ sin x / ( 1 + cos x ) ] = lim ( x -- > 0 ) tan ( x / 2 ) = tan 0 = 0
Hmmm we could have used limited development and find that when x->0 this expression is equal to x/2->0 But I didn't think about doing it like this ! Thank you for the idea !
It is indeed, mate, but the video is someway targeting the learners In keeping with khan's site, L'Hopital's rule is way ahead. www.khanacademy.org/math/ap-calculus-ab/ab-derivative-rules/modal/a/proving-the-derivatives-of-sinx-and-cosx (as an example; therefor the beginners wldnt get this video right here). Peace pal
The proof of this is required in the proof of the derivative of sin(x), in which case it would be contradictory, as we would be trying to use a fact that we are trying to prove. I’m guessing this is why this method is being used. Same goes for the limit of sin(x)/x :)
+Maor Yevdayev He can't because the proof of the derivative of cos(x) already relies on this limit so using L'Hospital's on this is just like assuming that this limit is equaled to zero blindly and then relying on that assume to proof that this limit is equaled to zero. Pretty mind blowing and false. So yeah that's why you can't derive the top and the bottom to prove this limit. Because you can't take derivative without evaluating it.
+Suraj Jimmy Then the limit equals 1/2, use L'Hospital's rule for the limit you want and you will get the limit of (sinx/2x) as x-->0 and this is equaled to 1/2 the limit of (sinx/x) as x-->0 which is equaled to 1/2 because the limit of (sinx/x) as x-->0 equals 1.
now you may ask how sin^2 x + cos^2 = 1 see sin x and cos x = op/hyp and adj/hyp respectively now square both of them we get op^2+adj^2/hyp^2 [by using the pythagorean theorem we can get op^2 + adj^2 = hyp^2] now hyp^2/hyp^2=1 hence proved
![Proofs: Lim sinx/x =1 and lim [cosx -1] /x =0 as x goes to zero from geometry](http://i.ytimg.com/vi/W0jiodfVGx0/mqdefault.jpg)
sin x / ( 1 + cos x ) = tan ( x / 2 )
so you can also evaluate it as : lim ( x -- > 0 ) sin x / x * lim ( x -- > 0 ) [ sin x / ( 1 + cos x ) ]
= lim ( x -- > 0 ) tan ( x / 2 )
= tan 0
= 0
but that's just being fancy to be honest, still awesome though
Nice crosshair, mind sharing the settings? Trying to improve my aim, thanks
W
Helpful man ❤
very good explain, thank you
Thank You
Thank you for the video!
That exactly is the definition of the derivative at point 0 of cos(x)
lim cos(0) - cos(x) / 0 - x = d/dx cos(0) = sin(0) = 0
x->0
Great explaination ,thanks!!
Love it!
mr. khan, i literally owe you my degree
Hmmm we could have used limited development and find that when x->0 this expression is equal to x/2->0
But I didn't think about doing it like this ! Thank you for the idea !
You rock Sal
is it possible to deduce that the limit as pi/2 - theta goes to zero of cos x/x is equal to 1 ?
lim(x>infinity) [sin x÷(x + cos x)] =?
Is the product of limits always equal to the limit of products?
yes
@@ruthyves392 Yes, assuming the limits exist. In this case they do, so it's safe to write an equality.
Salam alekum
You could use L'Hopital's rule instead, it is way easier
It is indeed, mate, but the video is someway targeting the learners
In keeping with khan's site, L'Hopital's rule is way ahead.
www.khanacademy.org/math/ap-calculus-ab/ab-derivative-rules/modal/a/proving-the-derivatives-of-sinx-and-cosx (as an example; therefor the beginners wldnt get this video right here). Peace pal
The proof of this is required in the proof of the derivative of sin(x), in which case it would be contradictory, as we would be trying to use a fact that we are trying to prove. I’m guessing this is why this method is being used. Same goes for the limit of sin(x)/x :)
+Maor Yevdayev
He can't because the proof of the derivative of cos(x) already relies on this limit so using L'Hospital's on this is just like assuming that this limit is equaled to zero blindly and then relying on that assume to proof that this limit is equaled to zero. Pretty mind blowing and false. So yeah that's why you can't derive the top and the bottom to prove this limit. Because you can't take derivative without evaluating it.
What about if denominator is x^2
+Suraj Jimmy
Then the limit equals 1/2, use L'Hospital's rule for the limit you want and you will get the limit of (sinx/2x) as x-->0 and this is equaled to 1/2 the limit of (sinx/x) as x-->0 which is equaled to 1/2 because the limit of (sinx/x) as x-->0 equals 1.
@@someone229 -1/2
So the limit when x tends to 0 of 1+ cos x : x is also 0?
No
The limit doesn't exist
pls can somebody tell me why 1- cos^2 x = sin^2 x
sin^2 x + cos^2 = 1
hence sin^2 x = 1 - cos^2 x
just a simple algebra
now you may ask how sin^2 x + cos^2 = 1
see
sin x and cos x = op/hyp and adj/hyp respectively
now square both of them we get op^2+adj^2/hyp^2 [by using the pythagorean theorem we can get op^2 + adj^2 = hyp^2]
now hyp^2/hyp^2=1
hence proved
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So what?