How to find $\lim_{x\to0}\frac{x-x\cos x}{\tan^3x}$ without using L'hopital rule or Taylor series? I found that exercise. However I can not find the answer. I know I can do it with L'hopital. However is there a way to solve it without using L'hopital nor Taylor series? If so show me please. I will appreciate it a lot. If there's no way, which solution seems easier? L'hopital?Taylor?
Hints: You know that $$\frac{\tan x}{x} \xrightarrow[x\to 0]{} 1$$ (as this is $\tan'(0)$) so $\frac{x}{\tan x} \xrightarrow[x\to 0]{} 1$.
You know that $\cos^2 x \xrightarrow[x\to 0]{} 1$.
And you know that
$$
\frac{1-\cos x}{\sin^2 x}=\frac{1-\cos x}{1-\cos^2 x} = \frac{1-\cos x}{(1-\cos x)(1+\cos x)} = \frac{1}{1+\cos x}\xrightarrow[x\to 0]{} \frac{1}{2}.
$$

Can you piece all of it together?
We know that
$\lim_{x\to 0}\frac{\sin x}{x} = 1$, which implies that $$\lim_{x\to 0}\frac{x}{\sin x} = 1$$

Now lets consider
 $$\begin{array}{lll}
\lim_{x\to 0} \displaystyle\frac{x-x\cos x}{\tan^3 x}&amp;=&amp;\lim_{x\to 0} \displaystyle\frac{x(1-\cos x)}{\tan x \tan^2 x}\\
&amp;=&amp;\lim_{x\to 0} \displaystyle\frac{x\cos x(1-\cos x)}{\sin x \tan^2 x}\\
&amp;=&amp;\bigg(\lim_{x\to 0} \displaystyle\frac{x}{\sin x }\bigg)\bigg(\lim_{x\to 0} \displaystyle\frac{\cos x(1-\cos x)}{ \tan^2 x}\bigg)\\
&amp;=&amp;1\cdot \lim_{x\to 0} \displaystyle\frac{\cos x(1-\cos x)}{ \tan^2 x}\\
&amp;=&amp;\lim_{x\to 0} \displaystyle\frac{\cos^3 x(1-\cos x)}{ \sin^2 x}\\
\end{array}$$

Can you take it from here?