Question

${lim}_{x\to 0}\frac{1-cos4\theta }{1-cos6\theta }$

Answer 1

${lim}_{x\to 0}\frac{1-cos4\theta }{1-cos6\theta }$

$={lim}_{0\to 0}\frac{2si{n}^{2}2\theta }{2si{n}^{2}3\theta }={lim}_{0\to 0}\frac{(sin2\theta {)}^2}{(sin3\theta {)}^2}$

$=\frac{{lim}_{0\to 0}{\left(\frac{sin2\theta }{2\theta }\right)}^2\times 4{\theta }^2}{{lim}_{0\to 0}{\left(\frac{sin3\theta }{3\theta }\right)}^2\times 9{\theta }^2}=\frac{1\times 4{\theta }^2}{1\times 9{\theta }^2}=\frac{4}{9}$