Question

${\displaystyle \underset{x\to 0}{lim}\frac{1-\cos 4\theta }{1-\cos 6\theta }}$ is ______.

पर्याय

• ${\displaystyle \frac{4}{9}}$

• ${\displaystyle \frac{1}{2}}$

• ${\displaystyle \frac{-1}{2}}$

• –1

Answer 1

${\displaystyle \underset{x\to 0}{lim}\frac{1-\cos 4\theta }{1-\cos 6\theta }}$ is ${\displaystyle \frac{4}{9}}$.

Explanation:

Given ${\displaystyle \underset{\theta \to 0}{lim}\frac{1-\cos 4\theta }{1-\cos 6\theta }}$

= ${\displaystyle \underset{\theta \to 0}{lim}\frac{2{\sin }^{2}2\theta }{2{\sin }^{2}3\theta }}$  .....${\displaystyle [\because 1-\cos \theta =2{\sin }^2 \frac{\theta }{2}]}$

= ${\displaystyle \underset{\theta \to 0}{lim}\frac{{\sin }^{2}2\theta }{{\sin }^{2}3\theta }}$

= ${\displaystyle \underset{\theta \to 0}{lim}{\left[\frac{\sin 2\theta }{\sin 3\theta }\right]}^2}$

= ${\displaystyle \underset{\begin{array}{c}\theta \to 0\\ 2\theta \to 0\\ 3\theta \to 0\end{array}}{lim}{\left[\frac{\frac{\sin 2\theta }{2\theta }\times 2\theta }{\frac{\sin 3\theta }{3\theta }\times 3\theta }\right]}^2}$

= ${\displaystyle {\left[\frac{2\theta }{3\theta }\right]}^2}$

= ${\displaystyle {\left(\frac{2}{3}\right)}^2}$

= ${\displaystyle \frac{4}{9}}$