Question

Lim x tends to 0 1-cos2x/3x^2

Answer 1

Answer:

The answer is $\frac{2}{3}$

Step-by-step explanation:

   $\displaystyle \lim_{x\to 0}\frac{1-\cos 2x}{3x^2}$

  $=\displaystyle \lim_{x\to 0}\frac{2\sin^2x}{3x^2}$, since $1-\cos 2\theta=2\sin^2\theta$

 $=\frac{2}{3}\displaystyle \lim_{x\to 0}(\frac{\sin x}{x})^2$

$=\frac{2}{3}(1)^2=\frac{2}{3}$, since $\displaystyle \lim_{x\to 0}\frac{\sin x}{x}=1$