Question

Calculati: $\lim_{x \to 0} \frac{1-cos \ x}{x^2}$ Fara L'H. Multumesc!

Answer 1

$\textbf{Folosesc limita remarcabil\u{a}}:$

$\lim\limits_{x\to a}\dfrac{\sin\Big(u(x)\Big)}{u(x)} = 1,\,\,\,\, \text{dac\u{a} }u(x)\to 0\\ \\ \\ \textbf{Metoda 1:}$

$\cos 2x = 1-2\sin^2 x \\ \Rightarrow 2\sin^2 x = 1 - \cos 2x\\ \Rightarrow 2\sin^2 \left(\frac{x}{2}\right) = 1-\cos x\\ \\\lim\limits_{x\to 0}\dfrac{1-\cos x}{x^2} =\lim\limits_{x\to 0}\dfrac{2\sin^2\left(\frac{x}{2}\right)}{x^2} = \lim\limits_{x\to 0}\left(\dfrac{2\sin^2\left(\frac{x}{2}\right)}{\frac{x^2}{4}} \cdot \dfrac{1}{4}\right) =$

$=2\lim\limits_{x\to 0}\dfrac{\sin^2\left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)^2}\cdot \dfrac{1}{4} = 2\lim\limits_{x\to 0}\left(\dfrac{\sin \left(\frac{x}{2}\right)}{\frac{x}{2}}\right)^2\cdot \dfrac{1}{4} =$

$= 2\cdot 1^2\cdot \dfrac{1}{4} = \boxed{\dfrac{1}{2}}\\ \\ \\ \textbf{Metoda 2:}$

$\lim\limits_{x\to 0}\dfrac{1-\cos x}{x^2} = \lim\limits_{x\to 0}\dfrac{1-\cos x}{x^2}\cdot 1 =$

$= \lim\limits_{x\to 0}\dfrac{1-\cos x}{x^2}\cdot \lim\limits_{x\to 0}\dfrac{1+\cos x}{2} =$

$=\lim\limits_{x\to 0}\left(\dfrac{1-\cos x}{x^2}\cdot \dfrac{1+\cos x}{2}\right)=$

$=\lim\limits_{x\to 0}\dfrac{1-\cos^2 x}{x^2\cdot 2} =\dfrac{1}{2}\lim\limits_{x\to 0}\dfrac{\sin^2 x}{x^2} =$

$=\dfrac{1}{2}\lim\limits_{x\to 0}\left(\dfrac{\sin x}{x}\right)^2= \dfrac{1}{2}\cdot 1^2 = \boxed{\dfrac{1}{2}}$