Question

Lim 1-cos^2 5x/3x^2,где x-> 0 с пределом!

Answer 1

$\lim _{x\to \:0}\left(\frac{1-\cos ^2\left(5x\right)}{3x^2}\right)=\frac{1}{3}\cdot \lim \:_{x\to \:0}\left(\frac{1-\cos ^2\left(5x\right)}{x^2}\right)= \\ \\ =\frac{1}{3}\cdot \lim \:_{x\to \:0}\left(\frac{5\sin \left(10x\right)}{2x}\right)=\frac{1}{3}\cdot \lim \:_{x\to \:0}\left(\frac{50\cos \left(10x\right)}{2}\right)= \\ \\ =\frac{1}{3}\cdot \lim \:_{x\to \:0}\left(25\cos \left(10x\right)\right)=\frac{1}{3}\cdot \:25\cos \left(10\cdot \:0\right)=\frac{25}{3} \\ \\ OTVET: \frac{25}{3}$

Answer 2

$\lim\limits _{x \to 0} \frac{1-cos^25x}{3x^2} = \lim\limits _{x \to 0} \frac{sin^25x}{3x^2}= \lim\limits _{x \to 0}\Big (\underbrace {\frac{sin5x}{5x}}_{1}\cdot \underbrace {\frac{sin5x}{5x}}_{1}\cdot \frac{25x^2}{3x^2}\Big )=\\\\= \lim\limits _{x \to 0} \frac{25x^2}{3x^2} = \frac{25}{3}$