# $\lim_{x\to0} \frac{x-\sin x}{x-\tan x}$ without using L'Hopital

$$\lim_{x\to0} \frac{x-\sin x}{x-\tan x}=?$$

I tried using $$\lim\limits_{x\to0} \frac{\sin x}x=1$$.

But it doesn't work :/

• I just don't know why people say solve *** without L'Hopital. Sep 29 '13 at 11:33
• Either the instructor doesn't want them to use l'H, or else they haven't yet studied it... Sep 29 '13 at 11:34
• @TedShifrin I tried divide it with x all the way but the result is 0/0 Sep 29 '13 at 11:50
• No, Sami's solution is not the best either. Using Taylor is just the same as using L'Hospital.
– OR.
Sep 29 '13 at 11:53
• @ABC: I disagree with you ardently. An experienced student can do the Taylor polynomial calculation in his head here; I challenge most people to do the product rule (and often chain rule) three or more times in succession in both numerator and denominator and not make any mistakes. The Taylor polynomial is both operationally and conceptually more appropriate and more indicative of what is actually going on. Sep 29 '13 at 12:25

In the beginning of this answer, it is shown that \begin{align} \frac{\color{#C00000}{\sin(2x)-2\sin(x)}}{\color{#00A000}{\tan(2x)-2\tan(x)}} &=\underbrace{\color{#C00000}{2\sin(x)(\cos(x)-1)}\vphantom{\frac{\tan^2(x)}{\tan^2(x)}}}\underbrace{\frac{\color{#00A000}{1-\tan^2(x)}}{\color{#00A000}{2\tan^3(x)}}}\\ &=\hphantom{\sin}\frac{-2\sin^3(x)}{\cos(x)+1}\hphantom{\sin}\frac{\cos(x)\cos(2x)}{2\sin^3(x)}\\ &=-\frac{\cos(x)\cos(2x)}{\cos(x)+1}\tag{1} \end{align} Therefore, $$\lim_{x\to0}\,\frac{\sin(x)-2\sin(x/2)}{\tan(x)-2\tan(x/2)}=-\frac12\tag{2}$$ Thus, given an $\epsilon\gt0$, we can find a $\delta\gt0$ so that if $|x|\le\delta$ $$\left|\,\frac{\sin(x)-2\sin(x/2)}{\tan(x)-2\tan(x/2)}+\frac12\,\right|\le\epsilon\tag{3}$$ Because $\,\displaystyle\lim_{x\to0}\frac{\sin(x)}{x}=\lim_{x\to0}\frac{\tan(x)}{x}=1$, which are shown geometrically in this answer, we have $$\sin(x)-x=\sum_{k=0}^\infty2^k\sin(x/2^k)-2^{k+1}\sin(x/2^{k+1})\tag{4}$$ and $$\tan(x)-x=\sum_{k=0}^\infty2^k\tan(x/2^k)-2^{k+1}\tan(x/2^{k+1})\tag{5}$$ By $(3)$ each term of $(4)$ is between $-\frac12-\epsilon$ and $-\frac12+\epsilon$ of the corresponding term of $(5)$.

Therefore, if $|x|\le\delta$ $$\left|\,\frac{\sin(x)-x}{\tan(x)-x}+\frac12\,\right|\le\epsilon\tag{6}$$ We can restate $(6)$ as $$\lim_{x\to0}\frac{x-\sin(x)}{x-\tan(x)}=-\frac12\tag{7}$$

• Telescope the partial sums of the series. You use the limits to get the convergence of the partial sums.
– OR.
Sep 29 '13 at 18:03
• @experimentX: for $(4)$ $$\sum_{k=0}^{n-1}2^k\sin(x/2^k)-2^{k+1}\sin(x/2^{k+1})=\sin(x)-2^n\sin(x/2^n)$$ and $$\lim_{n\to\infty}2^n\sin(x/2^n)=x$$ Similar for $(5)$, just replace $\sin$ with $\tan$.
– robjohn
Sep 29 '13 at 21:13
• If $\tan(x)<0$ and we have $$-\varepsilon-\frac12\le\frac{2^k\sin\left(\frac{x}{2^k}\right)-2\cdot 2^k\sin\left(\frac{\frac{x}{2^k}}2\right)}{2^k\tan\left(\frac{x}{2^k}\right)-2\cdot 2^k\tan\left(\frac{\frac{x}{2^k}}2\right)}\le\varepsilon-\frac12$$ shouldn't we say each term of $(4)$ is between $-\frac12-\varepsilon$ and $-\frac12+\varepsilon$ of the absolute value of the corresponding term of $(5)$? Dec 1 '20 at 20:54
• @Invisible: When the bottom is negative, the top is positive and vice-versa. The ratio is close to $-\frac12$ when $|x|$ is small, positive or negative (both numerator and denominator are odd functions).
– robjohn
Dec 1 '20 at 22:04

$$\frac{x - \sin(x)}{x - \tan(x)} = \frac{x - \sin(x)}{x^3} \cdot \frac{x^3}{x - \tan(x)}$$

Let $x = 3y$ and $x\to 0 \implies y\to 0$ $$\lim_{x\to0} \frac{x - \sin(x)}{x^3} = L$$ $$L = \lim_{y\to0}\frac{3y - \sin(3y)}{(3y)^3} = \lim_{y\to0} \frac 3 {27} \frac{y - \sin(y)}{y^3} + \lim_{y\to0} \frac{4}{27} \frac{\sin^3(y)}{y^3} = \frac{1}{9} L + \frac 4{27}$$

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

\begin{align*} L &= \lim_{y\to0}\frac{ 3y - \tan(3y)}{27 y^3} \\ &= \lim_{y\to0} \frac{1}{(1 - 3\tan^2(y ))} \cdot \frac{3y(1 - 3\tan^2(y )) - 3 \tan(y) + \tan^3(y)}{27y^3}\\ &= \lim_{y\to0} \frac{1}{(1 - 3\tan^2(y ))} \cdot \left( \frac 3 {27} \frac{y - \tan(y)}{y^3} + \frac 1 {27} \frac{\tan^3(y)}{y^3} - \frac 9 {27} \frac{y \tan^2(y)}{y^3 } \right )\\ &= \frac {3L}{27} + \frac 1 {27} - \frac 1 3 \\ \end{align*}

This gives other limit to be $-1/3$, put it up and get your limit.

• I wonder if you assume implicitly that $L$ exists. Sep 29 '13 at 12:54
• @Siminore yes that I assumed :( I can't find any other way. I saw that one for sine in yahoo answer couple of months ago ... but I can't seem to find it. Sep 29 '13 at 12:54
• @experimentX: this is very cute, assuming the limit exists. (+1). Proving the limit exists is a bit more difficult.
– robjohn
Sep 29 '13 at 16:56
• @robjohn thanks, but i had seen it done somewhere on yahoo answers. Sep 29 '13 at 16:58

Hint

Use the Taylor series: $$\sin x=x-\frac{x^3}{6}+o(x^3)\quad \text{and}\quad \tan x=x+\frac{x^3}{3}+o(x^3)$$

• Didn't learn about this yet. I have no idea how to use it... Sep 29 '13 at 12:17
• That was my initial reaction, @IndyZa. I assumed your class was at the beginning of the first course. I do not know what your teacher expects you to do. I suggest you inquire and let us know :) Sep 29 '13 at 12:21

$$L=\lim_{x\to0} \frac{x-\sin x}{x-\tan x}=\lim_{x\to0}\frac{x-\sin x}{x\cos x-\sin x}\cos x = \lim_{x\to0}\frac{2x-\sin2x}{2x\cos2x-\sin2x}\cos2x\\ = \lim_{x\to0}\frac{x-\cos x\sin x}{x(1-2\sin^2x)-\cos x\sin x}\cos2x=\lim_{x\to0}\frac{x-\cos x\sin x}{x-\cos x\sin x-2x\sin^2x}\cos2x$$ Which, noting that $\lim_{x\to0}\cos2x=1$, we can then write as $$\lim_{x\to0}\frac{1}{1-\frac{2x\sin^2x}{x-\cos x\sin x}} = \frac{1}{1-2\lim_{x\to0}\frac{x\sin^2x}{x-\cos x\sin x}}=\frac{1}{1-2M}$$ Now, we turn our attention to that new limit... $$\frac1M=\lim_{x\to0}\frac{x-\cos x\sin x}{x\sin^2x}=\lim_{x\to0}\frac{1-\cos x\frac{\sin x}x}{1-\cos^2x}=1+\lim_{x\to0}\frac{1-\frac{\sin x}{x\cos x}}{1-\cos^2x}\cos^2 x\\ =1+\lim_{x\to0}\frac{x-\tan x}{x\sin^2x}$$ But we also have $$\frac1M = \lim_{x\to 0} \frac{2x-\sin2x}{2x\sin^2x}=2\lim_{x\to 0} \frac{2x-\sin2x}{2x(1-\cos2x)}=2\lim_{x\to0}\frac{x-\sin x}{x(1-\cos x)}\\ =2\lim_{x\to0}\frac{x-\sin x}{x(1-\cos^2 x)}(1+\cos x)=4\lim_{x\to0}\frac{x-\sin x}{x(1-\cos^2 x)}=4\lim_{x\to0}\frac{x-\sin x}{x\sin^2x}\\ =4\lim_{x\to0}\frac{x-\sin x}{x-\tan x}\cdot\frac{x-\tan x}{x\sin^2x}$$ So, we have $$\frac1M = 4L\left(\frac1M-1\right)$$ or $1=4L(1-M)$... but $L=\frac{1}{1-2M}$ (or $1=L(1-2M)$).

Therefore, we have that $$1-2=4L-4LM-2L+4LM = 2L = -1$$ Therefore, $L=-\frac12$. No use of $\lim_{x\to0}\frac{\sin x}x$ required.

• +1. Quite a tour de force that I would never expect a beginning calculus student to do. Sep 29 '13 at 15:13
• This assumes that $\lim\lim_{x\to0}\frac{x\sin^2(x)}{x-\cos(x)\sin(x)}$ exists. Otherwise, it works. (+1)
– robjohn
Sep 29 '13 at 17:14
• @TedShifrin: Oh, I agree. I just noticed the potential to find it without needing $\frac{\sin x}x$, and figured I'd provide it. To be honest, I can't see a solution approach that one would expect a beginning calculus student to do. Sep 30 '13 at 1:12
• @robjohn: As with experimentX's solution, I've assumed the existence of the limits (for sake of conciseness). Sep 30 '13 at 1:20

Lemma $$\bf1$$: Suppose that both $$\prod\limits_{k=1}^\infty(1-a_k)$$ and $$\sum\limits_{k=1}^\infty a_k$$ converge. Then $$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to\infty}\frac{1-\prod\limits_{k=n}^\infty(1-a_k)}{\sum\limits_{k=n}^\infty a_k}=1}\tag1$$ Proof: \begin{align} \lim_{n\to\infty}\frac{1-\prod\limits_{k=n}^\infty(1-a_k)}{\sum\limits_{k=n}^\infty a_k} &=\lim_{n\to\infty}\frac{\prod\limits_{k=n+1}^\infty(1-a_k)-\prod\limits_{k=n}^\infty(1-a_k)}{\sum\limits_{k=n}^\infty a_k-\sum\limits_{k=n+1}^\infty a_k}\tag{1a}\\ &=\lim_{n\to\infty}\frac{a_n\prod\limits_{k=n+1}^\infty(1-a_k)}{a_n}\tag{1b}\\[15pt] &=\lim_{n\to\infty}\prod\limits_{k=n+1}^\infty(1-a_k)\tag{1c}\\[18pt] &=1\tag{1c} \end{align} Explanation:
$$\text{(1a)}$$: Stolz-Cesàro
$$\text{(1b)}$$: simplify numerator and denominator
$$\text{(1c)}$$: cancel numerator and denominator
$$\text{(1d)}$$: the product converges

$$\large\square$$

Lemma $$\bf2$$: Suppose that both $$\prod\limits_{k=1}^\infty(1+a_k)$$ and $$\sum\limits_{k=1}^\infty a_k$$ converge. Then $$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to\infty}\frac{1-\prod\limits_{k=n}^\infty(1+a_k)^{-1}}{\sum\limits_{k=n}^\infty a_k}=1}\tag2$$ Proof: \begin{align} \lim_{n\to\infty}\frac{1-\prod\limits_{k=n}^\infty(1+a_k)^{-1}}{\sum\limits_{k=n}^\infty a_k} &=\lim_{n\to\infty}\frac{\prod\limits_{k=n+1}^\infty(1+a_k)^{-1}-\prod\limits_{k=n}^\infty(1+a_k)^{-1}}{\sum\limits_{k=n}^\infty a_k-\sum\limits_{k=n+1}^\infty a_k}\tag{2a}\\ &=\lim_{n\to\infty}\frac{a_n\prod\limits_{k=n}^\infty(1+a_k)^{-1}}{a_n}\tag{2b}\\[15pt] &=\lim_{n\to\infty}\prod\limits_{k=n}^\infty(1+a_k)^{-1}\tag{2c}\\[18pt] &=1\tag2 \end{align} Explanation:
$$\text{(2a)}$$: Stolz-Cesàro
$$\text{(2b)}$$: simplify numerator and denominator
$$\text{(2c)}$$: cancel numerator and denominator
$$\text{(2d)}$$: the product converges

$$\large\square$$

Induction, $$\lim\limits_{x\to0}\frac{\sin(x)}x=1$$, and $$\frac{\sin^2(x)}{4\sin^2(x/2)}=\left(1-\sin^2(x/2)\right)$$ prove

Lemma $$\bf{3}$$: $$\bbox[5px,border:2px solid #C0A000]{\frac{\sin^2(x)}{x^2}=\prod_{k=1}^\infty\left(1-\sin^2\left(x/2^k\right)\right)}\tag3$$

Induction, $$\lim\limits_{x\to0}\frac{\tan(x)}x=1$$, and $$\frac{\tan(x)}{2\tan(x/2)}=\left(1-\tan^2(x/2)\right)^{-1}$$ prove

Lemma $$\bf{4}$$: $$\bbox[5px,border:2px solid #C0A000]{\frac{\tan(x)}{x}=\prod_{k=1}^\infty\left(1-\tan^2\left(x/2^k\right)\right)^{-1}}\tag4$$

The Limit \begin{align} \lim_{x\to0}\frac{x-\sin(x)}{x-\tan(x)} &=\frac12\lim_{x\to0}\left(1+\frac{\sin(x)}{x}\right)\frac{1-\frac{\sin(x)}{x}}{1-\frac{\tan(x)}x}\tag{5a}\\ &=\frac12\lim_{x\to0}\frac{1-\frac{\sin^2(x)}{x^2}}{1-\frac{\tan(x)}x}\tag{5b}\\ &=\frac12\lim_{n\to\infty}\frac{1-\frac{\sin^2\left(x/2^n\right)}{\left(x/2^n\right)^2}}{1-\frac{\tan\left(x/2^n\right)}{x/2^n}}\tag{5c}\\ &=\frac12\lim_{n\to\infty}\frac{1-\prod\limits_{k=n+1}^\infty\left(1-\sin^2\left(x/2^k\right)\right)}{1-\prod\limits_{k=n+1}^\infty\left(1-\tan^2\left(x/2^k\right)\right)^{-1}}\tag{5d}\\ &=-\frac12\lim_{n\to\infty}\frac{\sum\limits_{k=n+1}^\infty\sin^2\left(x/2^k\right)}{\sum\limits_{k=n+1}^\infty\tan^2\left(x/2^k\right)}\tag{5e}\\[6pt] &=-\frac12\lim_{n\to\infty}\frac{\sin^2\left(x/2^n\right)}{\tan^2\left(x/2^n\right)}\tag{5f}\\[18pt] &=-\frac12\tag{5g} \end{align} Explanation:
$$\text{(5a)}$$: $$\lim\limits_{x\to0}\frac12\left(1+\frac{\sin(x)}x\right)=1$$
$$\text{(5b)}$$: algebra
$$\text{(5c)}$$: rewrite the limit
$$\text{(5d)}$$: Lemma $$3$$ and Lemma $$4$$
$$\text{(5e)}$$: Lemma $$1$$ and Lemma $$2$$
$$\text{(5f)}$$: Stolz-Cesàro
$$\text{(5g)}$$: $$\lim\limits_{x\to0}\cos^2(x)=1$$