Question

Limit from $x$ to $0$ of $\frac{\left[1-\cos \left(1-\cos \left(x\right)\right)\right]}{x^4}=$

Answer 1

Compute the required value:

Given: $\underset{x\to 0}{\lim }\frac{\left[1-\cos \left(1-\cos \left(x\right)\right)\right]}{x^4}$

That is

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{\left[1-\cos \left(1-\cos \left(x\right)\right)\right]}{x^4} \\ =\underset{x\to 0}{\lim }\frac{\left[1+2{\sin }^2\left({\displaystyle \frac{1-\cos \left(x\right)}{2}}\right)-1\right]}{x^4}\left[\because cos\left(2x\right)=1-2si{n}^2\left(x\right)\right] \\ =\underset{x\to 0}{\lim }\frac{\left[2{\sin }^2\left({\displaystyle \frac{1+2{\sin }^2\left({\displaystyle \frac{x}{2}}\right)-1}{2}}\right)\right]}{x^4} \\ \underset{x\to 0}{=\lim }\frac{\left[2{\sin }^2\left({\displaystyle \frac{2{\sin }^2\left({\displaystyle \frac{x}{2}}\right)}{2}}\right)\right]}{x^4} \\ =2\times \underset{x\to 0}{\lim }\frac{{\sin }^2\left({\displaystyle {\sin }^2\left(\frac{{\displaystyle x}}{{\displaystyle 2}}\right)}\right)}{{\sin }^4\left(\frac{{\displaystyle x}}{{\displaystyle 2}}\right)}\times \frac{{\sin }^4\left(\frac{{\displaystyle x}}{{\displaystyle 2}}\right)}{{\left({\displaystyle \frac{x}{2}}\right)}^4}\times {\left(\frac{1}{2}\right)}^4 \\ =2\times \underset{x\to 0}{\lim }{\left(\frac{\sin \left({\displaystyle {\sin }^2\left(\frac{{\displaystyle x}}{{\displaystyle 2}}\right)}\right)}{{\sin }^2\left(\frac{{\displaystyle x}}{{\displaystyle 2}}\right)}\right)}^2\times \frac{{\sin }^4\left(\frac{{\displaystyle x}}{{\displaystyle 2}}\right)}{{\left({\displaystyle \frac{x}{2}}\right)}^4}\times {\left(\frac{1}{2}\right)}^4\left[as\underset{x\to 0}{lim}\frac{sin\left(x\right)}{x}=1\right] \\ =2\times \frac{1}{2^4}\times 1\times 1 \\ =\frac{1}{2^3} \\ =\frac{1}{8} \end{gathered}$$

Hence, the value of $\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{\left[1-\cos \left(1-\cos \left(x\right)\right)\right]}{{\mathbf{x}}^{\mathbf{4}}}$ is $\frac{1}{8}$.