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

Solution 1:

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 $$ \lim_{x\to0}\frac{x-\sin(x)}{x-\tan(x)}=-\frac12\tag{7} $$

Solution 2:

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

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.

Solution 3:

Hint

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

Solution 4:

$$ L=\lim_{x\to0} \frac{x-\sin x}{x-\tan x}=\lim_{x\to0}\frac{x-\sin x}{x\cos x-\sin x}\cos x = \lim_{x\to0}\frac{2x-\sin2x}{2x\cos2x-\sin2x}\cos2x\\ = \lim_{x\to0}\frac{x-\cos x\sin x}{x(1-2\sin^2x)-\cos x\sin x}\cos2x=\lim_{x\to0}\frac{x-\cos x\sin x}{x-\cos x\sin x-2x\sin^2x}\cos2x $$ 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.

Solution 5:

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
$\text{(5g)}$: $\lim\limits_{x\to0}\cos^2(x)=1$