Okay, here it is. First, remember that: sin^2(y) = (1/2)(1-cos(2y)) (cos(h)-1) / h = - (1-cos(h)) / h = - 2sin^2(h/2) / h = - sin(h/2)*sin(h/2) / (h/2) Just for simplicity, to better see things. Let's call "p = h/2". As h ---> 0, so does p, multiplying by a half here is irrelevant since everything is scaled accordingly. This gives us lim(p ---> 0): - sin(p)*sin(p)/p Since the last sin(p)/p = 1 when p goes to zero, that part is known, but we also multiply that with something else, namely sin(p), which goes to zero. So it is essentially - 0 * 1= 0. And there you go. DO NOT USE TAYLOR EXPANSION TO SOLVE THIS! The Taylor expansion is based on the fact that one already knows what the derivatives for sin(x) and cos(x) are (because that is how their expansions are derived). So what you do then is trying to use a tool whose properties are built on derivatives. Essentially trying to prove sin(x)'s derivative, by using sin(x)'s derivative. It is a circle argument. The same goes for l'Hospital's rule. Use figures, trigonometric properties, known limits or whatever. Using Taylor or l'Hospital's rule requires knowledge when they can be applied. These are common mistakes on exams.
You're right, it is technically an indeterminate form, and can't be evaluated the way he did it. You can't use L'Hopital's rule either, since it requires differentiation, which jumps the gun. A better non-rigorous way to evaluate the limits would be to use the small angle approximation that cos(x)=1-x^2/2 and sin(x)=x for x
That is not okay to do. Because the Taylor expansion of cos(x) and sin(x) builds on that one knows their derivatives. So by doing that you would use their derivatives to prove their derivatives, and that circle argument is forbidden. He doesn't do it correctly here entirely, because he doesn't show that cos(h)-1 part goes faster to zero than the sin(h) part.
+modemanslutning You (as well as others) are right! Eddie Woo missed a (small) part here. (but remember: he's not teaching university students here ;-)) He should have shown that (cos(x)-1)/x → 0 as x → 0. (merely stating that cos(x)-1 → 0 is not enough, indeed) 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 additional 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), of course. (which we should not use at this point, since we know neither the derivative of sin(x), nor that of cos(x) yet) Now, using the fact that sin'(x) = cos(x), we could derive that cos'(x) = -sin(x) either by using trig identities for cos(x+Δx) (and the limit above), or by using the fact that cos(x) = sin (π/2 - x), together with the chain rule.
+Navjot Saroa I'm afraid I have to disagree with your statement above, on two grounds even. Firstly, he is not - as you state - taking the limit of an expression where the numerator is identically 0. Instead, he is taking the limit as h approaches 0 of an expression of the form f(h)/g(h) where both f(h) as well as g(h) are approaching 0. (actually g(h)=h here). Secondly, even if f(h)→0 as h→0, this is no guarantee for the limit of the expression f(h)/h to even exist as h→0, let alone it being 0. If we take e.g. f(x)=x^α, then lim (h→0+) f(h)/h only exists if α≥1, otherwise it approaches +∞. (if α is irrational, the value of f(x) is not defined for x
sir 1st of all your way of teaching is amazing and i m a big fan of yours . now about this video , i want u to break lim x-o [sin(x+h)-sin(x)]/h in sina-sinb formula :- [sin(x+h)-sinx]/h =[ 2cos(x+h+x/2).sin(x+h-x/2)]/h < .symbolises multiplication sign> =[2cos(2x+h/2).sin(h/2)]/h now divide and multiply sin(h)/2 by h/2 . then we get :- =[2cos(2x+h/2) . h/2]/h now , this h/2 divided by h gives 1/2 and = 1/2 . 2cos(2x+h/2) gives ;- cos(2x+h/2) with limit x-0 we get cos(2x/2) = cos(x)
7:55 - BIG MISTAKE there! Yes, limit of cos(h) is approaching 1 as h approaches 0, BUT how can you just neglect the h in the denominator here? This is a big shortcut, with no sensible explanation.
You're right, it is an indeterminate form and the way he evaluated that was totally wrong. It would have been better, although not totally rigorous, to use the small angle approximations for sine and cosine.
You basically have to prove that cos(h) - 1 approaches 0 FASTER than h approaches 0. This isn't a rigorous proof, sure, but if you graph y=cos(x)-1, you can kind of see why it approaches 0 faster than y = x. The line approaches (0,0) from a much higher position than y = x and approaches 0 at a very low gradient, which sorta demonstrates why it's a lot closer to 0. Though the two equations intersect at (0,0), cos(x)-1 has obviously been CLOSER to 0 for the whole time while approaching (0,0).
for those who are confuse with cos h - 1/h = 0/0 actually its not 0/0 ^^ see Lets forget about that denominator h and focus on cos h -1 okay expand it ( cos h -1 = ( 1- sin^2(h/2) -1) = sin^2(h/2) = ( sin(h/2) x sin(h/2) ) ) i guess u know this now in sin(y) when y is very small its value = y example sin(1/10000000000000) equivalent to 1/10000000000000 so sin(h/2) [as h -> 0 (very small ) ] = h/2 so that cos h-1/h will become (h/2 x h/2)/h cancel h and h u will get h/4 put limit u will get 0 i hope its clear to you now ^^ its not 1-1 =0 you just cant forget same variable one over another
What's your point? It was still 0/0 in the form that was in the video and the way he evaluated it was still wrong. No one disagrees that the limit was equal to 0, they just have issue with the way that he evaluated it.
For those of you puzzled, to really prove the limit of 1-cosh/h and sinh/h, you should really NOT do this or memorize what he did, but examine other sources (google Khan academy has a very long winded explanation but its better than making the grand assumptions he did).
So much energy! That was really nice, but I was really hoping you'd go through the Lim h-> 0 sin(h)/h = 1 part. Can you say that sin(h) = h since they both equal zero as h ->0 ?
Muhammad Abdur Rahman as we know that h tends to 0 and don't apply h value in the denominator in the beginning itself! First apply in the numerator and then we get cos0=1 and 1-1/h=0/h=0 and we know that something divided by 0 is 0 itself and hence the solution!
actually its not 0/0 ^^ see Lets forget about that denominator h and focus on cos h -1 okay expand it ( cos h -1 = (1- sin^2(h/2) -1) = sin^2(h/2) = ( sin(h/2) x sin(h/2) ) ) i guess u know this now in sin(y) when y is very small its value = y example sin(1/10000000000000) equivalent to 1/10000000000000 so sin(h/2) [as h -> 0 (very small ) ] = h/2 so that cos h-1/h will become (h/2 x h/2)/h cancel h and h u will get h/4 put limit u will get 0 i hope its clear to you now ^^ its not 1-1 =0 you just cant forget same variable one over another
haru shizuku seperate the limits Lim as h tends to 0 cos h - 1 × lim as h tends to 0 1/h Now apply the value for the first limit you get 0 now as we multiply 0 with anything no matter what the other number is it'll definitely be zero ;) there's nothing wrong in your solution too, but I generally use easy methods to solve problems so I used the seperation of limits and we can apply the limit value individually to the same variables when we seperate them.
@@thepositionalplay5307 You can't separate limits like that, if you apply that exact same argument to the sin(x)/x example, you get 0, and that is WRONG. It's called indeterminent form, its a very well documented calculus 2 topic.
I'm an indian. Do you think we need to look into a sheet to prove the derivative? 2nd last step is crucial which is not dealt clearly looking at the watch
@@alang.2054 L'H's rule would be circular reasoning. You use the squeeze theorem to bound the value of sin(h)/h between 1 and tan(h), in order to show that this has to be true from first principles.
Sir can you please show me the graphical representation of derivative of sin x. Derivative mean change in x wrt y. How graph of sin x change to cos x graph. Please explain this.
Sketch a graph of sin(x), and its key features at every multiple of pi/2 for x. At x = 0, sin(x) is rising at its maximum slope, and passing thru the origin. This implies its derivative is maximum and positive. At x=pi/2, sin(x) is flat, and at its maximum value. This implies its derivative is zero. At x=pi, sin(x) is descending at its extreme negative slope. This implies its derivative is negative and is minimum. At x=3*pi/2, sin(x) is flat, and at its minimum value. This implies its derivative is zero. At x=2*pi, it returns to its rising node at the maximum slope, where its derivative is maximum. Follow only what the slope does in the above. Max positive : zero : min negative : zero : back to max positive. Now notice cos(x)'s behavior: x=0, cos(x) is maximum and positive x=pi/2, cos(x) is zero at x=pi, cos is minimum and negative x=3*pi/2, cos(x) is zero Notice that cos(x) has the same key points as each first derivative of sin(x)? That's how sin's derivative visually relates to cosine.
This could have been done without the factoring by sin x and just splitting the numerator into three fractions over h. And from sketching a right angle triangle it is easy to to see that (cos h)/h --> 1 as h --> 0.
Though the result is correct, the method is logically flawed. When we start getting 0/0 we know we are in trouble - we need to use l'hopital or the squeeze theorem
