How do I evaluate this limit (without the use of L'Hospital or series expansions)?

Question

I have been trying for quite some time to evaluate this limit

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

but without using L'Hospital's rule or series expansions. I have tried the common things, I have got this:

$$\lim_{x\to0} \frac{\sin x(1-\cos x)}{\cos x(x-\sin x)}$$

But I have no idea how to proceed further.

Answer 1

Hint: Here the key limit will be derived. The rest should be easy.

Let: $$ L=\lim_{x\to0}\frac{x-\sin x}{x^3}. $$ We have: $$ L=\lim_{x\to0}\frac{2x-\sin 2x}{8x^3}\implies 4L=\lim_{x\to0}\frac{x-\frac12\sin 2x}{x^3}\\\implies 3L=\lim_{x\to0}\frac{\sin x-\frac12\sin 2x}{x^3} =\lim_{x\to0}\frac{\sin x(1-\cos x)}{x^3} =\lim_{x\to0}\frac{\sin^3 x}{x^3(1+\cos x)}=\frac12\\ \implies L=\frac16. $$

Answer 2

We will be using $\lim_{z \to 0} \frac{\sin z}{z}=1$ frequently. $$L=\lim_{x\to 0} \frac{1}{\cos x} \lim_{x\to 0} \frac{\sin x}{x}\lim_{x \to 0}\frac{\frac{1}{2}\sin^2(x/2)}{(x/2)^2} \lim_{x\to 0} \frac{x^3}{x-\sin x} =L/2~~~~(1)$$ Let $L=\frac{1}{A}$, where $$A=\lim_{x\to 0} \frac{x-\sin x}{x^3}= \lim_{x\to 0} \frac{3x-\sin 3x}{(3x)^3}=\lim_{x\to 0} \frac{3x -3\sin x +4 \sin^3 x}{27x^3}.$$ $$\implies A=A/9+\frac{4}{27} \implies \frac{8A}{9}=\frac{4}{27} \implies A=\frac{1}{6}~~~~(2).$$ Using (2) in (1) we get the required limit $L=3.$