Question

If $\underset{x\to 0}{lim}\left\{\frac{1}{x^8}\left(1-\cos \frac{x^2}{2}-\cos \frac{x^2}{4}+\cos \frac{x^2}{2}\cos \frac{x^2}{4}\right)\right\}=2^{-k}$, then the value of $k$ is:

Answer 1

Step 1: Simplify $\underset{x\to 0}{lim}\left\{\frac{1}{x^8}\left(1-\cos \frac{x^2}{2}-\cos \frac{x^2}{4}+\cos \frac{x^2}{2}\cos \frac{x^2}{4}\right)\right\}$

$\begin{array}{rcl}\underset{x\to 0}{lim}\left\{\frac{1}{x^8}\left(1-\cos \frac{x^2}{2}-\cos \frac{x^2}{4}+\cos \frac{x^2}{2}\cos \frac{x^2}{4}\right)\right\}& =& \underset{x\to 0}{lim}\left[\frac{1}{x^8}\left\{1-\cos \frac{x^2}{2}-\cos \frac{x^2}{4}\left(1-\cos \frac{x^2}{2}\right)\right\}\right]\\ & =& \underset{x\to 0}{lim}\left[\frac{1}{x^8}\left\{\left(1-\cos \frac{x^2}{2}\right)\left(1-\cos \frac{x^2}{4}\right)\right\}\right]\\ & =& \underset{x\to 0}{lim}\left[\frac{1}{x^8}\left(2{\sin }^2\frac{x^2}{4}\right)\left(2{\sin }^2\frac{x^2}{8}\right)\right]\left[by1-\cos 2\theta =2{\sin }^2\theta \right]\\ & =& \underset{x\to 0}{lim}\left[\frac{4}{x^4·x^4}{\left(\sin \frac{x^2}{4}\right)}^2{\left(\sin \frac{x^2}{8}\right)}^2\right]\\ & =& \underset{x\to 0}{lim}\left[4\frac{{\left(\sin \frac{x^2}{4}\right)}^2}{{\displaystyle \frac{16x^4}{16}}}·\frac{{\left(\sin \frac{x^2}{8}\right)}^2}{{\displaystyle \frac{64x^4}{64}}}\right]\\ & =& \underset{x\to 0}{lim}\left[4\frac{1}{{\displaystyle 16}}·\frac{1}{{\displaystyle 64}}\right]\left[by\underset{x\to 0}{lim}\frac{\sin x}{x}=1\right]\\ & & \\ & =& \underset{x\to 0}{lim}\left[\frac{1}{{\displaystyle 4}}·\frac{1}{{\displaystyle 64}}\right]\\ & =& \underset{x\to 0}{lim}\left[\frac{1}{{\displaystyle 256}}\right]\end{array}$

Step 2: Find the value of $k$.

We have $\underset{x\to 0}{lim}\left\{\frac{1}{x^8}\left(1-\cos \frac{x^2}{2}-\cos \frac{x^2}{4}+\cos \frac{x^2}{2}\cos \frac{x^2}{4}\right)\right\}=2^{-k}$

$\begin{array}{rcl}& \Rightarrow & \frac{1}{256}=2^{-k}\\ \Rightarrow \frac{1}{2^8}& =& 2^{-k}\\ \Rightarrow 2^{-8}& =& 2^{-k}\\ \Rightarrow k& =& 8\end{array}$

Therefore, $k=8$