Question

$\underset{x\to 0}{lim}\frac{8}{x^8}[1-cos\frac{x^2}{2}-cos\frac{x^2}{4}+cos\frac{x^2}{2}cos\frac{x^2}{4}]\text{is equal to}$

• A. 1/16
• B. -1/16
• C. 1/32
• D. -1/32

Answer 1

The correct option is C

1/32

$=\underset{x\to 0}{lim}\frac{8}{x^8}[1-cos\frac{x^2}{2}-cos\frac{x^2}{4}+cos\frac{x^2}{2}cos\frac{x^2}{4}]=\underset{x\to 0}{lim}\frac{8}{x^8}[(1-cos\frac{x^2}{2})-cos\frac{x^4}{4}(1-cos\frac{x^2}{2})]=\underset{x\to 0}{lim}\frac{8}{x^8}(1-cos\frac{x^2}{2})(1-cos\frac{x^4}{4})=\underset{x\to 0}{lim}\frac{8}{x^8}.2si{n}^2\frac{x^2}{4}.2si{n}^2\frac{x^2}{8}=\underset{x\to 0}{lim}\frac{32}{x^8}\left(\frac{sin\frac{x^2}{4}}{\frac{x^2}{4}}\right)\left(\frac{x^2}{4}\right)\left(\frac{sin\frac{x^2}{8}}{\frac{x^2}{8}}\right){\left(\frac{x^2}{8}\right)}^2=\frac{1}{32}$