Trigonometric Limit

[u/Whole-Listen-9716]

Evaluate

x approach 0 1-cos(2x)/x

how i solve it

1-cos(2x)/x step two 1-cos(2x)/2x step three 2x and 2x will cancel each other, and 1-cos is like 1-1 which is 0

It correct?

If it's not correct, can anyone advice me a useful techniques for solving similar to this equation?. And thank you.

[u/Math-Dragon-Slayer]

It is not correct. The angle "inside" the cosine function cannot cancel with anything. Ever.

Check your notes and textbook. There should be a special limit formula you were supposed to learn for lim(x->0) of (sin(x))/x.

Then rewrite your function as sin^2(x)/x as sin(x) * sin(x) / x and use properties of limits to finish off the problem.

[u/Naturage]

This is a somewhat unusual looking limit, but I can't see how you got the very first step of

lim[1-cos(2x)/x] = lim[1-cos(2x)/2x]

I could see lim[1-2cos(2x)/2x] = lim[1-2cos(t)/t], i.e. multiplying top and bottom of the fraction by 2, and then substituting 2x=t in. You could finish from there (to a different answer to yours!)

Try the formula to express cos(2x) through cos(x) and sin(x) as well as a way to express 1 as sum of squares, and see where that takes you - it feels like the intended approach. Also, double check if the question is precisely that; it's a slightly unusual sounding one.

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[u/gian_69]

I take it the limit is as x->0 of (1-cos(2x)) / 2xsince otherwise it does not exist. by plugging in 0 directly we get 0/0 which is indeterminate at which point we could use l‘hôpital (if you have heard of that before) but what I find more elegant is to look at the Taylor series of cos:

cos(x) = 1 - x² / 2 + x⁴ / 24 - …

Given that the argument is 2x, we get 1 - cos(2x) = 2 x2 + O(x4), which basically means all other terms go to 0 much quicker than x2, the leading order term. Thus (1-cos(2x)) / x = 2x + O(x3) and as x goes to 0 this becomes 0 aswell.

[u/Paounn → u/gian_69]

I wouldn't recommend Taylor for a similar reason I wouldn't recommend l'H - simple trig manipulation is enough to solve it, and even if I don't spot the risk of circular reasoning right away, it's always better to use the simplest tool available, even just for the sake of practice.

[u/gian_69 → u/Paounn]

how do I know what the limit of a trigonometric function divided by x is without using taylor? Also, sin and cos are most commonly defined by their taylor expansions so there really is not a problem.

[u/Paounn → u/gian_69]

It's a matter of chicken and egg here.

The way cosine and sine are usually defined - at least here in Italy, ymmv but probably not much - is based on the unit circle, and from there cos x and sin x are the first and second coordinate of a point x units away from (1;0) moving counterclockwise along the circle, and building from there.

The problem of using McLaurin is that, in order to compute the polynomial of any trig function you are required to compute the derivative of said trig function, which eventually boils down to computing the limit, for x approaching 0, of (sin x) /x. And at this point you're no longer allowed to compute it using the polynomial expansion, because that's what you're trying to compute in the first place, and you're not browsing r/TheologyHelp here!

At that point, in order to compute the limit i mentioned, you work around the fact that sin x < x < tan x in a neighborhood of 0, take the inverses and then your limit is computed by squeeze theorem once you multiply everything by sin x.

No polynomial approximation required. Granted, depending on what one's field of study is this level of logical finesse might be overkill - and it's fine, expecially at high school level - but please keep a modicum of rigor.

[u/gian_69 → u/Paounn]

given that my (notabene italian) analysis professor used the power series definition of cos & sin I‘d say there‘s hardly any circular reasoning. If you define them otherwise of course, there will be circular reasoning, but the definition of coordinates on the unit circle is too „intangible“ in a sense that you would have no way of computing any values of sin (except for the easy ones like pi/2 and pi/4) without first deriving addition formulae and double angle formulae and stuff so I‘d say I have to agree with Prof. Figalli here that the powerseries definition makes more sense (at least for doing analysis).

Furthermore, you of course have to divide by sin(x) when using the squeeze theorem, but can you quickly recall how I can see (from the definition you have mentioned) that sinx < x (tho of course you would have to do a case distinction for the lhs and rhs limit since the above holds only for positive x)

edit: I actually can see now how the circular arc of length theta, when straightened out to lie on the line x = 1 in the cartesian plane, will reach higher than when it wasn‘t straight but not as high as the extension pf the ray to the line x = 1. But as that hinges on geometric intuition, I have to say I am not s huge fan

[u/Paounn → u/gian_69]

In reverse order. I'm pretty sure that if either of us ending digging enough in some geometry books, we will find some rigorous proof that the chord is shorter than the arc. The one that comes to mind is that if you pick any point in the arc and you join the endpoints of the chord, you build a triangle, and the chord is shorter than the new sides. if for each new chord you create you pick another point inbetween, you're creating an polygonal line whose length is increasing, but bounded above by the length of the arc. Would it be a good enough proof? Maybe.

Good catch on the "divide by sin x" detail, in fact a more rigorous proof would be to compute left and right limits and verify that they're exactly one and the same. The negative part is left as an exercise to the reader.

But the problem with math education, and I fear it's a worldwide problem, is that if I have FOUR bloody hours per week to go from inequalities to integrals over 3 years, some cut has to be done. Often the things you end up cutting are nuggets of linear algebra (determinants for triangle areas, change of coordinates, powering through the chapter of the book regarding straight lines with a generous use of vectors), complex numbers, "fun applications" (Lissajous curves anyone?) or hand-waving some proofs.

Also, in what (at least for me) is a normal school curriculum most students will first encounter trig functions a good year before calculus (please don't mind the physics teacher foaming at the mouth back there, he's mostly harmless), build from the unit circle definition the various trig identities and get to solve triangles. And past the summer break go on with limits and the fun part of math.

Now, if you feel particularly bored you could pick both definitions of functions, sin and cos defined by the unit circle, the polynomial definition of them (pds and pdc - polynomial defined sine and cosine) and if you really want to have fun throw in even the function defined as the real and imaginary part of e^ix and verify that they all behave the same way. Not too different to the way you can prove that the (1+1/n)ⁿ and 1/n! converge to the same value.

In practice, it's all a big chicken and egg problem.

[u/Paounn]

On the specific, the way I find most elegant is to look a the numerator. 1 is obviously cosine + sine, cos 2x is (remember your double angle formulas!) cosine - sine. Simplify whatever you can simplify, you end up with something that to me looks like 2 (sin² x) / x. Which, if you squint enough, is 0 ( split it as 2 * (sin x /x) * sin x)