Question

lim x tends to 0 (1-cos^3(2x))/x^2cos4x

Answer 1

Solution:

$\lim_{x \to 0}\frac {(1-cos^3(2x))}{x^2 cos4x}\\\\ \lim_{x \to 0}\frac{(1-cos 2x)(1+cos^2 2x+Cos2x)}{x^2\times cos4x}\\\\ \lim_{x \to 0}\frac{2 sin^2x \times(1+1+1)}{x^2\times 1}\\\\ \lim_{x \to 0}\frac{sin^2x}{x^2}\times (2 \times 3})\\\\ 1\times 2\times 3=6\\\\ \lim_{x \to 0}\frac{sinx}{x}=1, so ,[\lim_{x \to 0}(\frac{sinx}{x})^2]=1$

Used these mathematical concepts while solving this problem

$a^3 -b^3=(a-b)(a^2+ab+b^2)\\\\ \lim_{x \to 0}cos 2x=1,  \\\\ \lim_{x \to 0}cos 4x=1,$

cos 2 x= 2 cos²x-1

           = 1 - 2 sin² x