# Evaluate $\lim\limits_{x\to 0} \frac{x(1-\cos x)}{x - \sin(x)}$ without Taylor series or L'Hôpital's rule?

I want to evaluate $$\lim\limits_{x\to 0} \frac{x(1-\cos x)}{x - \sin(x)}$$. We know that its plot is:

And also with attention to the Taylor series we know that its series expansion at $$x=0$$ is:

$$3 - x^2/10 - x^4/4200 + O(x^6)$$

So the limit is $$3$$. But I want to evaluate it without Taylor series or L'Hôpital's rule.

What I tried: \begin{aligned} \lim\limits_{x\to 0} \frac{x(1-\cos x)}{x - \sin(x)}& =\\ &\lim_{x\to 0}\frac{x\cdot2\sin^2\frac{x}{2}}{x - \sin x} = \lim_{x\to 0}\frac{2x\cdot\sin^2\frac{x}{2}}{x\cdot(1 - \frac{\sin x}{x})} = \\&\lim_{x\to 0}\frac{2\sin^2\frac{x}{2}}{1 - \frac{\sin x}{x}} \end{aligned} But that did not help to evaluate the indeterminate form. Also I want to know is there a geometric representation for $$x - \sin(x)$$ which helps us to evaluate the limit?

• An immediate thought that comes to mind is that $$\lim\limits_{x\to 0} \frac{x(1-\cos x)}{x - \sin(x)} = \lim\limits_{x\to 0} \frac{1-\cos x}{x} \cdot \frac{x}{1 - \frac 1 x \sin x}$$ The first factor has a special limit, not sure about the second factor at the moment. Commented Jul 16, 2023 at 4:03
• You might be interested in this question, which comes from your problem by multiplying by $\frac{\sin x}{x}$ (which amounts to multiplying by $1$ in the limit) Commented Jul 16, 2023 at 4:05
• Commented Jul 16, 2023 at 4:16
• It's trivial with a Taylor polynomial. So what you are trying is nonsrnse Commented Jul 17, 2023 at 8:00
• @gnasher729 Even if a problem already has a known and explored trivial way in mathematics, a new way of approaching the problem can still have value. It can help foster creativity, critical thinking, and a deeper understanding of mathematical concepts and methods, and potentially lead to improved or alternative solutions. Commented Jul 17, 2023 at 9:25

Use two standard limits:

(1) $$\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac{1}{2}$$

(2) $$\lim_{x\to0}\frac{x-\sin x}{x^3}=\frac{1}{6}$$

So your question is nothing but ratio of two:

Proof of (1) $$\lim_{x\to0}\frac{1-\cos x}{x^2}=\lim_{x\to0}\frac{2\sin^2(\frac{x}{2})}{x^2}=\frac{1}{2}\lim_{x\to0}(\frac{\sin(\frac{x}{2})}{\frac{x}{2}})^2=\frac{1}{2}$$

Proof of (2)

$$L=\lim_{x\to0}\frac{x-\sin x}{x^3}$$ $$\Rightarrow L=\lim_{x\to0}\frac{3x-\sin(3x)}{(3x)^3}$$

$$\Rightarrow L=\lim_{x\to0}\frac{3x-3\sin x+4 \sin^3 x}{27x^3}$$ $$\Rightarrow L=\frac{L}{9}+\frac{4}{27}\lim_{x\to0}(\frac{\sin x}{x})^3$$ $$\Rightarrow \frac{8L}{9}=\frac{4}{27}$$ $$\Rightarrow L=\frac{1}{6}$$

• please correct your last 3 lines; so you will get "$L=\frac{1}{6}$". Commented Jul 16, 2023 at 5:02
• The given argument does not `prove' (2). It assumes the existence of the limit, but it may not exist(although it exists in this case). To prove without L'Hopital's rule or Taylor series, one may use mean value theorem third or four times. Commented Jul 16, 2023 at 5:08
• @Riemann, Can you explain your idea about "mean value theorem" or provide a link to that? Commented Jul 16, 2023 at 6:02

Too long for a comment: An Idea for the geometric meaning question of OP.

In freshman level calculus we proved the limits $$\lim_{x\to 0}\frac{\sin x}x=1\tag 0$$ and $$\lim_{x\to 0}\frac{1-\cos x}{x^2}=\frac12\tag1$$ geometrically by using the sector $$\widehat{OPA}$$ of the unit circle where $$O$$ is the origin, $$P=(\cos x,\sin x)$$ and $$A=(1,0)$$. The area of this sector minus the area of the triangle $$\triangle OPA$$ is equal to $$\frac{x-\sin x}2$$. We can approximate this difference area by drawing equally spaced $$n$$ chords (or tangent lines) on the arc and constructing a polygon. The resulting expression will be a function of $$\cos(\frac x{2n})$$ and $$\sin(\frac x{2n})$$ in chord way. Dividing by $$x^3$$ and taking the limit $$x\to 0$$ and $$n\to\infty$$ we can find the limit $$\frac{1}{12}$$ by using $$(0)$$ and $$(1)$$ only. Indeed, I did this with three tangent lines and obtained $$\frac{3}{32}$$ which is an upper bound for $$\frac{1}{12}$$. Hence, $$\lim_{x\to 0}\frac{x-\sin x}{x^3}=\frac16\tag2$$ The answer of the question:

$$\lim\limits_{x\to 0} \frac{x(1-\cos x)}{x - \sin x}=\lim\limits_{x\to 0} \frac{\frac{1-\cos x}{x^2}}{\frac{x - \sin x}{x^3}}=\frac{\frac12}{\frac16}=3.$$

Addition to the answer of @deep dm. We know that $$\sin' = \cos$$ and $$\cos' = -\sin$$.

Let $$f(x) = \sin x-x$$. Since $$f(0) = 0$$ and $$f'(x) = \cos(x)-1\le 0$$ for $$x\in[0, \frac{\pi}{2}]$$, by MVT $$f(x)\le 0$$ for $$x\in[0, \frac{\pi}{2}]$$.

Similarly, let $$g(x) = \cos(x)-1+\frac{x^2}{2}$$. $$g(0) = 0$$ and $$g'(x) = -\sin x+x = -f(x)\ge 0$$ for $$x\in[0,\frac{\pi}{2}]$$, by MVT we have $$g(x)\ge 0$$ for $$x\in[0, \frac{\pi}{2}]$$.

Again, let $$h(x) = \sin(x)-x+\frac{x^3}{6}$$. $$h(0) = 0$$ and $$h'(x) = g(x)\ge 0$$ for $$x\in[0, \frac{\pi}{2}]$$ and $$h(x)\ge 0$$ for $$x\in[0, \frac{\pi}{2}]$$.

Doing this stuff two times more, one obtains $$x-\frac{x^3}{6}\le\sin x\le x-\frac{x^3}{6}+\frac{x^5}{24}$$. This is enough to conclude that the limit (2) is $$\frac{1}{6}$$.

It may seem dificult to find right coefficient(actually, they are Taylor coefficients). One may use integration(define $$g(x) = -\int_0^x f(t)dt$$, etc.) instead of MVT, then the coefficients arise naturally, too.