Compute $\lim_{x\to 0} \frac{x-\sin x}{x-\tan x}$ with L'hopital rule. Compute $\lim_{x\to 0} \frac{x-\sin x}{x-\tan x}$ with L'hopital rule.
Well, I know that I need to use the Quotient rule first $(\frac{f}{g})'$. but then, what do I have to do next?
Edit: I meant by quotient rule: $\frac{f'g - fg'} {g^2}$
 A: By successive application of L'hopital rule twice,
$$\lim_{x\to 0} \frac{x-\sin x}{x-\tan x}$$
$$=\lim_{x\to 0} \frac{1-\cos x}{1-\sec^2 x}$$
$$=\lim_{x\to 0} \frac{\sin x}{-2\sec^2 x \tan x}$$
$$=-\frac{1}{2}\lim_{x\to 0} \cos^3 x$$
$$=-\frac{1}{2}$$
A: Note that while we are living near zero, we get : $$x-\sin x\approx x^3/6-x^5/120, x-\tan x\approx  -x^3/3-2x^5/15$$
A: $$\lim_{x\to 0}\frac{x-\sin(x)}{x-\tan(x)}=\lim_{x\to 0}\frac{\frac{\text{d}}{\text{d}x}\left(x-\sin(x)\right)}{\frac{\text{d}}{\text{d}x}\left(x-\tan(x)\right)}=\lim_{x\to 0}\frac{1-\cos(x)}{1-\sec^2(x)}=$$
$$-\lim_{x\to 0}\frac{\cos^2(x)}{1+\cos(x)}=-\frac{\cos^2(0)}{1+\cos(0)}=-\frac{1}{2}$$
A: We will prove this without the use of L'Hospital's rule. Let $L= \lim_{x \to 0} \dfrac{x-\sin(x)}{x-\tan(x)}$. We then have
$$L = \lim_{x \to 0} \dfrac{x-\sin(x)}x \cdot \dfrac{x}{x-\tan(x)}$$ Assume $\lim_{x \to 0} \dfrac{x-\sin(x)}{x^3} = M$ and $\lim_{x \to 0} \dfrac{x-\tan(x)}{x^3} = N$ exists. We then have
$$M = \lim_{x \to 0} \dfrac{3x-\sin(3x)}{(3x)^3} = \lim_{x \to 0} \dfrac{3x-3\sin(x)+4\sin^3(x)}{27x^3} = \dfrac19\lim_{x \to 0} \dfrac{x-\sin(x)}{x^3} + \dfrac4{27}\lim_{x \to 0} \dfrac{\sin^3(x)}{x^3}$$
Hence, we obtain that
$$M = \dfrac{M}9 + \dfrac4{27} \implies \dfrac{8M}9 = \dfrac4{27} \implies M = \dfrac16$$
Similarly, we have
\begin{align}
N & = \lim_{x \to 0} \dfrac{3x-\tan(3x)}{(3x)^3} = \lim_{x \to 0} \dfrac{3x-\dfrac{3\tan(x)-\tan^3(x)}{1-3\tan^2(x)}}{27x^3} = \lim_{x \to 0} \dfrac{3x-3\tan(x)-9x\tan^2(x)+\tan^3(x)}{27x^3(1-3\tan^2(x))}\\
& = \dfrac19 \cdot \lim_{x \to 0}\dfrac{x-\tan(x)}{x^3} \cdot \dfrac1{1-3\tan^2(x)} + \lim_{x \to0}\left(\dfrac1{27}\cdot\dfrac{\tan^3(x)}{x^3} - \dfrac13 \cdot \dfrac{\tan^2(x)}{x^2}\right)\cdot \dfrac1{1-3\tan^2(x)}\\
& = \dfrac{N}9 + \dfrac1{27}-\dfrac13 = \dfrac{N}9 - \dfrac8{27}
\end{align}
Hence, we obtain that
$$N = \dfrac{N}9 - \dfrac8{27} \implies \dfrac{8N}9 = -\dfrac8{27} \implies N = -\dfrac1{3}$$
Hence, $$L = \dfrac{M}N = \dfrac{1/6}{-1/3} = - \dfrac12$$
