Question

plz solve this question I need it​

Answer 1

✪ᴀɴsᴡᴇʀ✪

• $\boxed{\bf{ \lim \limits_{x \to 0}\frac{8}{x^8}\left[1-cos\frac{x^2}{2}-cos\frac{x^2}{4}+cos\frac{x^2}{2}cos\frac{x^2}{4}\right] = \frac{1}{32}}}$

᯽ғᴏʀᴍᴜʟᴀ᯽

• $\lim \limits_{x \to 0} \frac{1-cos(x)}{x^2} = \frac{1}{2}$

⍟ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴ⍟

The given limit can be simplified as

❥︎ $\lim \limits_{x \to 0}\frac{8}{x^8}\left[1-cos\frac{x^2}{2}-cos\frac{x^2}{4}+cos\frac{x^2}{2}cos\frac{x^2}{4}\right]$

❥︎ $\lim \limits_{x \to 0}\frac{8}{x^8}\left[\left(1-cos\frac{x^2}{2}\right)-cos\frac{x^2}{4}\left(1-cos\frac{x^2}{2}\right)\right]$

❥︎ $\lim \limits_{x \to 0}\frac{8}{x^8} \left(1-cos\frac{x^2}{2}\right)\left(1-cos\frac{x^2}{4}\right)$

❥︎ $8\lim \limits_{x \to 0} \frac{1-cos\frac{x^2}{2}}{x^4}\frac{1-cos\frac{x^2}{4}}{x^4}$

❥︎ $8\lim \limits_{x \to 0} \frac{1-cos\frac{x^2}{2}}{2^2\left(\frac{x^2}{2}\right)^2}\frac{1-cos\frac{x^2}{4}}{4^2\left(\frac{x^2}{4}\right)^2}$

❥︎ $\frac{8}{4. 16}\lim \limits_{\frac{x^2}{2} \to 0} \frac{1-cos\frac{x^2}{2}}{\left(\frac{x^2}{2}\right)^2}\lim \limits_{\frac{x^2}{4} \to 0} \frac{1-cos\frac{x^2}{4}}{\left(\frac{x^2}{4}\right)^2}$

❥︎ $\frac{8}{4.16}\lim \limits_{y \to 0} \frac{1-cos(y)}{y^2}\lim \limits_{z \to 0} \frac{1-cos(z)}{z^2}$

❥︎ $\frac{1}{8}.\frac{1}{2}.\frac{1}{2}$

❥︎ $\frac{1}{32}$