TITLE: $\lim_{x\to0} \frac{x-\sin x}{x-\tan x}$ without using L'Hopital
QUESTION [9 upvotes]: $$\lim_{x\to0} \frac{x-\sin x}{x-\tan x}=?$$
I tried using $\lim\limits_{x\to0} \frac{\sin x}x=1$.
But it doesn't work :/

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