How to find $\lim_{x\to0}\frac{x-x\cos x}{\tan^3x}$ without using L'hopital rule or Taylor series?

Question

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?

Answer 1

First split off a factor $\frac x{\tan x}$ which has limit 1, and write the remaining trigonometric functions in terms of $\tan\frac x2.$

Answer 2

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?

Answer 3

We can write, whenever $\sin x\neq 0$ and $\cos x\neq -1$ \begin{align} \frac{x-x\cos x}{\tan^3 x}&=\frac{x(1-\cos x)}{\sin^3 x/\cos^3 x}\\ &=(\cos^3 x)\cdot \frac{x}{\sin x}\cdot\left(\frac{1-\cos x}{\sin^2 x}\right)\\ &=(\cos^3 x)\cdot \frac{1}{\frac{\sin x}{x}}\cdot\left(\frac{1-\cos x}{\sin^2 x}\cdot\frac{1+\cos x}{1+\cos x}\right)\\ &=\frac{\cos^3 x}{1+\cos x}\cdot \frac{1}{\frac{\sin x}{x}}\cdot\left(\frac{\sin^2 x}{\sin^2 x}\right)\\ &=\frac{\cos^3 x}{1+\cos x}\cdot \frac{1}{\frac{\sin x}{x}} \end{align} Then \begin{align} \lim_{x\to 0}\frac{x-x\cos x}{\tan^3 x}&=\left(\lim_{x\to 0}\frac{\cos^3 x}{1+\cos x}\right)\cdot \left(\frac{1}{\lim_{x\to 0}\frac{\sin x}{x}}\right)\\ &=\left(\frac{1}{1+1}\right)\cdot \left(\frac{1}{1}\right)\\ &=\color{blue}{\frac{1}{2}} \end{align}

Answer 4

Notice, $$\lim_{x\to 0}\frac{x-x\cos x}{\tan^3x}=\lim_{x\to 0}\frac{x(1-\cos x)}{\tan^3x}$$ $$=\lim_{x\to 0}\frac{x\left(2\sin^2\left(\frac{x}{2}\right)\right)}{\tan^3x}$$

$$=\frac{1}{2}\lim_{x\to 0}\frac{\left(\frac{\sin\left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)}\right)^2}{\left(\frac{\tan x}{ x}\right)^3}$$

$$=\frac{1}{2}\lim_{x\to 0}\left(\frac{\sin\left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)}\right)^2\cdot \lim_{x\to 0}\frac{1}{\left(\frac{\tan x}{ x}\right)^3}$$ $$=\frac{1}{2}(1)^2\cdot \frac{1}{(1)^3}=\color{red}{\frac12}$$

Answer 5

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

Can you take it from here?