$\lim_{x\to0} \frac{x-\sin x}{x-\tan x}$ 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 :/

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