# How do I evaluate $\lim _{x\to 0}\left(\frac{x-\sin x}{x\sin x}\right)$ without using L'Hopital or series?

How do I evaluate $$\lim _{x\to 0}\left(\frac{x-\sin x}{x\sin x}\right)$$ without using L'Hopital or series?

I've tried expanding the variable such as $$x = 2y$$ or $$x = 3y$$, but seemed to still get stuck.

Once we've proven $$\lim_{x\to0}\frac{\sin x}{x}=1$$ and $$\lim_{x\to0}\frac{x-\sin x}{x^3}=\frac16$$ (see e.g. here and here for solutions using neither L'Hôpital's rule nor series), your limit is $$\lim_{x\to0}\frac{x-\sin x}{x^3}\frac{x^2}{\sin x}=0$$.

First, read this answer. This answer shows that, for $$x$$ close to $$0$$, we have the following:

$$\sin x \leq x \leq \tan x$$

Thus, since $$\sin x \leq x$$ and $$\sin x$$ has the same sign as $$x$$ (i.e. either both are positive, both are negative, or both are $$0$$), we know that $$\frac{1}{\sin x} \geq \frac{1}{x}$$. Thus, $$\frac{1}{\sin x}-\frac{1}{x}\geq 0$$.

Also, since $$x \leq \tan x$$ and $$\tan x$$ has the same sign as $$x$$, we know that $$\frac{1}{x} \geq \frac{1}{\tan x}$$. This implies that $$\frac{1}{\sin x}-\frac{1}{x} \leq \frac{1}{\sin x}-\frac{1}{\tan x}$$,

Now, we have:

$$0 \leq \frac{1}{\sin x}-\frac{1}{x} \leq \frac{1}{\sin x}-\frac{1}{\tan x}$$

Simplify the trig expressions:

$$0 \leq \frac{x-\sin x}{x\sin x} \leq \frac{1-\cos x}{\sin x}$$

Now, $$\lim_{x\to 0} 0=0$$ for obvious reasons. Thus, let's consider $$\lim_{x\to 0} \frac{1-\cos x}{\sin x}$$. Multiply numerator and denominator by $$1+\cos x$$:

$$\lim_{x\to 0}\frac{1-\cos^2 x}{\sin x(1+\cos x)}=\lim_{x\to 0}\frac{\sin^2 x}{\sin x(1+\cos x)}=\lim_{x\to 0}\frac{\sin x}{1+\cos x}=\frac{\sin 0}{1+\cos 0}=0$$

Thus, $$\lim_{x\to 0} 0=\lim_{x\to 0} \frac{1-\cos x}{\sin x}=0$$, so by Squeeze Theorem, $$\lim_{x\to 0}\frac{x-\sin x}{x\sin x}=0$$.

Try the following

$$L=\lim \dfrac{x-x\cos\frac{x}{2}+x\cos\frac{x}{2}-\sin x}{x\sin x}$$ then use the $$\sin(x)=2\sin \frac{x}{2}\cos\frac{x}{2}$$ now the first two terms limit is zero and the second two terms have a relation with the limit $$L$$ $$\lim \frac{x \cos\frac{x}{2} -2 \sin\frac{x}{2} \cos\frac{x}{2}}{2x \sin\frac{x}{2} \cos\frac{x}{2}}$$

$$\lim \frac{2 \cos\frac{x}{2}\left( \frac{x}{2}- \sin\frac{x}{2}\right)}{2x \sin\frac{x}{2} \cos\frac{x}{2}}$$

$$\lim \frac{1}{2}\frac{\frac{x}{2}-\sin \frac{x}{2}}{\frac{x}{2}\sin\frac{x}{2}}$$ and the last limit is $$1/2L$$ so $$L=1/2L$$ hence $$L=0$$ by this method you can solve something like $$\frac{\sin x -x}{x^3}$$ but you need to add and subtract a certain term can figure it out ?