Ok, you say you need to show that as t → 0, \( y =sin\,t → t\) . You want to show that the limit isn't \( \frac {sin\,t}{0}\)
I don't think it's a particularly rigorous way to go about showing the limit exists. Basically you seem to be saying that as t → 0, so does y, so y/t → t/t. Is that right?
I don't think you can use similar triangles to show \(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\). But you can use geometric reasoning and a little bit of analysis. If you look at a right-triangle constructed inside an arc, you have points O (0,0), A (cos θ, 0), B(cos θ, sin θ) , C(1, 0) you can prove \(0 \lt \theta \leq \frac{\pi}{2} \Rightarrow \sin \theta \lt \sqrt{2 - 2 \cos \theta} \lt \theta\) where the middle term is the straight line between B and C while the right term is the curved arc between B and C with radius 1. So \(0 \lt \theta \leq \frac{\pi}{2} \Rightarrow \frac{\sin \theta}{\theta} \lt 1\). And if you look at an n-gon inscribed within a unit circle, each side has length \(2 \sin \theta = 2 \sin \frac{\pi}{n}\) and is subsumed by an arc of \(2 \theta = \frac{2 \pi}{n}\). So the area of 1/n of a n-gon must be \(\frac{1}{2} \cos \theta \times 2 \sin \theta = \cos \frac{\pi}{n} \sin \frac{\pi}{n} \) while the area of 1/n of a circle is \( \theta = \frac{\pi}{n}\). Therefore the ratio of the area a n-gon to the circle which circumscribes is must be \(\cos \theta \frac{ \sin \theta }{ \theta} = \frac{ n \cos \frac{\pi}{n} \sin \frac{\pi}{n} }{ \pi} \). In the limit of \(n\to\infty\) there is zero area between the n-gon and the circle, so \(\lim_{n\to\infty} \frac{ n \cos \frac{\pi}{n} \sin \frac{\pi}{n} }{ \pi} = 1 = \lim_{\theta\to 0} \cos \theta \frac{ \sin \theta }{ \theta} \) Since \(0 \lt \theta \leq \frac{\pi}{2} \Rightarrow \cos \theta < 1 \) means that \(0 \lt \theta \leq \frac{\pi}{2} \Rightarrow \cos \theta \frac{ \sin \theta }{ \theta} < \frac{ \sin \theta }{ \theta} < 1 \) And since the limit of the left and right terms are 1, the limit of the middle term must be one. —— Just using differential analysis, l'Hospital's rule allow you to write almost immediately write: \(\lim_{\theta\to 0} \frac{\sin \theta}{\theta} = \lim_{\theta\to 0} \frac{\partial_{\theta} \sin \theta}{\partial_{\theta} \theta} = \lim_{\theta\to 0} \frac{\cos \theta}{1} = \cos 0 = 1\) —— \(2 - 2 \cos \theta = 4 \sin^2 \frac{\theta}{2}\) Fun fact.
I already know that one. Means "Well, I've never seen that before, and it's not in my math book." Nice try though. No. I'm saying that I need to prove that as t → 0, y → t. I didn't say anything about y → 0. Don't put words in my mouth. It's very impolite. Now, if you don't agree that I need to prove that as t → 0, y → t, why not? You seem inarticulate on this point. The entire reason we use limits is to avoid precisely the situation where t = 0, because t is on the bottom of a fraction and that's a Bad Never Do Thing. So we say, "in the limit of t going to 0, y goes to t, and as y approaches t, y/t approaches 1." See how that works?
That is not rigorous, because in expression \(\lim_{t \to 0} y(t) \) if it can be evaluated, has to be a number. While t is not a number, but a bound variable. See https://en.wikipedia.org/wiki/Free_variables_and_bound_variables Therefore \(\lim_{t \to 0} y(t) = t\) is not sound. While \(\lim_{t \to 0} y(t) = 0\) doesn't tell you if \(\lim_{t \to 0} \frac{y(t)}{0} \) is zero, non-zero or doesn't exist. Examples of functions where \(\lim_{t \to 0}f(t) = 0\) but \(\lim_{t \to 0} \frac{f(t)}{0}\) either doesn't exist or is not one exist: \(\frac{-1}{\ln t}\), \(\sqrt{t}\), \(t^2\), \(2e^x - e^{-2x} - e^x\). Since \(\sin t = \frac{-i}{2} \left( e^{it} - e^{-it} \right)\) and \(e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!}\), it follows that \(\frac{\sin t}{t} = \sum_{k=0}^{\infty} \frac{(-1)^k t^{2k}}{(2k+1)!} = 1 - \frac{t^2}{6} + \frac{t^4}{120} - \dots\) which converges everywhere and lets you conclude quickly that \(\lim_{t \to 0} \frac{\sin t}{t} = 1\).
OK, so we have one expression for sin(t), that is sin(t) = y, and now we need an expression for t, so that we can get the bound variable completely out of sin(t)/t. Let's see now... How about arcsin(y)? Yes, on a unit circle, arcsin(y) = t by definition because sin(t) = y. And in fact even not on a unit circle. So sin(t)/t = y/arcsin(y), on any circle. And therefore, we are looking to prove that \(\lim_{t \to 0} \frac{y} {arcsin \,y} = 1\) Now let's plug that in. The first line is still correct, but must be amended: 1. sin(t) = y, and t = arcsin(y). Thus we have sin(t)/t = y/arcsin(y) 2. Therefore, to show that \(\lim_{t \to 0} \frac{sin \,t} {t} = 1\) we must show that \(\lim_{t \to 0} \frac{y} {arcsin \,y} = 1\) which we can do by showing that as arcsin(y) → 0, y → arcsin(y). 3. Now, arcsin(0) = 0 Thus, as y → 0, arcsin(y) → 0. And, of course, that implies that as arcsin(y) → 0, y → 0; both functions are dependent upon y, so that's allowed; it's not circular. 4. QED; we have shown that \(\lim_{t \to 0} \frac{y} {arcsin \,y} = 1\) and therefore that \(\lim_{t \to 0} \frac{sin \,t} {t} = 1\) Now, is that rigorous?
OR \(\lim_{y \to 0} \frac{y} {\sin^{-1} y} = 1\) Agreed. No. This shares the same second defect of your original scheme. \(\lim_{x \to 0} f(x) = 0\) and \(\lim_{x \to 0} g(x) = 0\) are not sufficient to prove that \(\lim_{x \to 0} \frac{f(x)}{g(x)} = 1\) as per my examples. \(\lim_{x \to 0} \frac{e^x - e^{-2x}}{2e^x - e^{-2x} - e^{-x}} = \frac{3}{5}\) \(\lim_{x \to 0} \frac{ \sqrt{1+x} - \sqrt{1-x} - \ln (x+1) }{x} = 0\) No. For reasons stated above. In geometry, it's asking for the ratio of the length of a straight line to a curved line. This requires a larger perspective, like analysis to demonstrate the equality. Look at my previous triangle ABC AB = sin θ = 2 sin (θ/2) cos (θ/2) AC = 1 - cos θ = 2 sin(θ/2) sin(θ/2) BC = 2 sin (θ/2) while the curve from B to C is still θ So 0 < θ < π/2 means sin θ < 2 sin θ/2 < 4 sin (θ/4) < 8 sin (θ/8) < ... < θ Asymptotic flatness of a small portion of a circle suggests that \(\lim_{n\to\infty} 2^n \sin \frac{\theta}{2^n} = \theta\) but you won't find that in the type of geometric argument that's been seen up to here. And that is enough to show that \(\lim_{n\to\infty} \frac{2^n}{\theta} \sin \frac{\theta}{2^n} = 1 \) (dividing finite by finite is legal) which is enough to show \(\lim_{t\to 0} \frac{\sin t}{t} = 1\).
Thanks for taking the time, rpenner. I'm learning a lot from this. More in a little bit, it's been a wild morning. I gotta eat.
Hmmm, but this is \(\lim_{x \to 0} g(y) = 0\) or in context \(\lim_{t \to 0} \frac {y}{arcsin \,y} = 1\) As far as I can see y is a free variable. There is no more t in there. Also, I am not trying to establish the equivalent of \(\lim_{x \to 0} \frac{f(x)}{g(x)} = 1\); I already established that \(\frac {sin \,t}{t} = \frac {y}{arcsin \,y}\) in a previous step. If this is true, then \(\lim_{t \to 0} \frac {sin \, t}{t} = \lim_{t \to 0} \frac {y}{arcsin \,y} = 1\). Did I make a mistake somewhere there?
Another fun fact is there doesn't seem to be a way to construct a solution to \( cos(x) = sinc(x) \). So far, I've found one value of x ≠ 0 by guesswork using a calculator. Which is: 1.4302π, accurate to 3 decimals. Right, poorly worded question. But the similar triangles argument does show the tangent exists and is larger than the arc length/subtended angle.
Nope. By definition \(y = \sin t\) so you are merely suppressing the dependency on bound variable t by using abbreviated notation. Actually, that is exactly what you need to do to solve it. The problem is Euclidean Geometry, of itself, doesn't give you the tools to do it. You need concepts of continuity and limits and perhaps for all of them, differential calculus. \(\lim_{t \to 0} \frac {y}{arcsin \,y} \) is more explicitly written as \(\lim_{t \to 0} \frac {y(t)}{\textrm{arcsin} \, y(t)} \) which is shorthand for \(\lim_{t \to 0} \frac {\sin t}{\textrm{arcsin} \, \sin t} \) or \(\lim_{t \to 0} \frac {\sin t}{t} \). Merely changing the notation doesn't change that you have a limit of a ratio where both numerator and denominator have 0 as the limiting value, as since this is not finite, you can't say \(\frac{0}{0} = 1\) for the reason that counter examples exist. Only if \(y = \sin t\) so this is change of variables. Better separately written as \(\lim_{t \to 0} \frac {\sin \, t}{t} = \lim_{y \to 0} \frac {y}{\textrm{arcsin} \,y}\) by change of variables and \(\lim_{y \to 0} \frac {y}{\textrm{arcsin} \,y} = 1\) by some formal argument which has not been given. Here is a geometric argument that \(\lim_{t \to 0} \frac {\sin \, t}{t} = 0\) adapted from http://www.themathpage.com/acalc/sine.htm Let us have 0 < θ < π/2. Let have a unit circle and the associated points: O=(0,0), A=(1, 0), B=(cos θ, sin θ), C=(cos θ, –sin θ), D=(cos θ, 0), E=(sec θ, 0). Then the arc from A to B has measure θ. And the arc from B through A to C has measure 2 θ. So it follows geometrically that BC < ∡BAC < BE + EC We have line segments: OA = OB = OC = 1 OD = cos θ OE = sec θ BD = CD = sin θ BE = CE = tan θ So we have 0 < θ < π/2 ⇒ 0 < 2 sin θ < 2 θ < 2 tan θ from geometry. Since all values are positive we are allowed to divide all terms by 2 sin θ to get: 0 < θ < π/2 ⇒ 0 < 1 < θ / sin θ < sec θ Since all values are positive we are allowed to take multiplicative inverse if we reverse the direction of comparison: 0 < θ < π/2 ⇒ cos 0 < sin θ / θ < 1 Finally since \(\lim_{\theta\to 0} \cos \theta = 1 \lim_{\theta\to 0} 1 \) So because sin θ / θ is sandwiched between them, \(\lim_{t \to 0} \frac{\sin t}{t} = 1\).
Awesome, thanks rpenner. I'll have to think about it a while longer. Look for something maybe tonight or maybe tomorrow but not in the morning.
You are looking for solutions of \(x \cos x - \sin x = 0\) which is a transcendental equation. You can polish approximate solutions. Let an approximate solutions be \(x_0(n) = \frac{\pi}{2} (2n + 1) \) and \(f(x) = x \cos x - \sin x \), and \(g(x) = x - \frac{f(x)}{f'(x)} = x + \frac{\cos x}{\sin x} - \frac{1}{x} \) Then \(x_0(n), \; g(x_0(n)), \; g(g(x_0(n))) , \; g(g(g(x_0(n)))), \dots\) form an ever-improving approximate solution. \(x_0(1) = \frac{3 \pi}{2} \approx 4.712 \\ x_1(1) = g(x_0(1)) = \frac{9 \pi^2 - 4}{6 \pi} \approx 4.50018 \\ x_2(1) = g(g(x_0(n))) = \frac{ 81 \pi^4 - 108 \pi^2 + 16 }{54 \pi^3 - 24 \pi} + \tan \frac{2}{3 \pi} \approx 4.4934195 \\ x_3(1) = g(g(g(x_0(n)))) \approx 4.493409457931699 ... \approx 4.49340945790906417530788092728 \)
Looking at the plot of sinc(x) + cos(x) (a construction), the intersections are solutions to a transcendental equation, so there is no rational solution (a set of points lying on a, well, rational curve)? That construction: Please Register or Log in to view the hidden image! . . . already has a map of a transcendental number in it, along the x axis, so the intersections \( \{ cos(x) \cap sinc(x)\} \) just have to be non-rational multiples of \( \pi \), i.e. [we can only calculate] rational approximations? Using your approximation method we have to take \( g(g(g ... (g(x_0)) ... )) \) infinitely, like a limit?
There is (0,1) but no others. The intersections are \( \{ (x,y) : x \in \mathbb{R} \wedge y = \cos x \} \cap \{ (x,y) : x \in \mathbb{R} - \{ 0 \} \wedge x y = \sin x \vee x = 0 \wedge y = 1 \} \) (0,1) is easy. \( x_2(n) = \frac{ (2 n + 1)^4 \pi^4- 12 (2 n + 1)^2 \pi^2 + 16 }{ 2 (2n + 1)^3 \pi^3 - 8 (2n +1 ) \pi } + \tan \frac{2}{(2n +1) \pi} \) leads to is pretty good approximations for \(n \in \mathbb{Z} - \{ -1, 0 \}\). Exactly like a limit. But except for n = -1 or n=0, it converges quickly.
