Question

Lim que tende a zero 1-cos/x2= 1/2

Answer 1

$\lim\limits_{x\to0}\dfrac{1-\cos\,x}{x^{2}}=\lim\limits_{x\to0}\dfrac{(1-\cos\,x)\cdot(1+\cos\,x)}{x^{2}\cdot(1+\cos\,x)}\\\\\\\lim\limits_{x\to0}\dfrac{1-\cos\,x}{x^{2}}=\lim\limits_{x\to0}\dfrac{1^{2}-\cos^{2}x}{x^{2}(1+\cos\,x)}\\\\\\\lim\limits_{x\to0}\dfrac{1-\cos\,x}{x^{2}}=\lim\limits_{x\to0}\dfrac{1-\cos^{2}x}{x^{2}\cdot(1+\cos\,x)}$

Pela relação fundamental da trigonometria, temos que

$\sin^{2}x+\cos^{2}x=1~\Leftrightarrow~1-\cos^{2}x=\sin^{2}x$

logo

$\lim\limits_{x\to0}\dfrac{1-\cos\,x}{x^{2}}=\lim\limits_{x\to0}\dfrac{\sin^{2}x}{x^{2}\cdot(1+\cos\,x)}\\\\\\\lim\limits_{x\to0}\dfrac{1-\cos\,x}{x^{2}}=\lim\limits_{x\to0}\bigg[\dfrac{\sin\,x}{x}\cdot\dfrac{\sin\,x}{x}\cdot\dfrac{1}{1+\cos\,x}\bigg]$

Note que $\lim\limits_{x\to0}\dfrac{\sin\,x}{x},~\lim\limits_{x\to0}\dfrac{1}{1+\cos\,x}$ existem, pois o primeiro é o limite fundamental e o segundo não apresenta indeterminações. Portanto, o limite do produto é o produto dos limites, e

$\lim\limits_{x\to0}\dfrac{1-\cos\,x}{x^{2}}=\bigg[\lim\limits_{x\to0}\dfrac{\sin\,x}{x}\bigg]\cdot\bigg[\lim\limits_{x\to0}\dfrac{\sin\,x}{x}\bigg]\cdot\bigg[\lim\limits_{x\to0}\dfrac{1}{1+\,cos\,x}\bigg]\\\\\\\lim\limits_{x\to0}\dfrac{1-\cos\,x}{x^{2}}=1\cdot1\cdot\dfrac{1}{1+\cos\,0}\\\\\\\lim\limits_{x\to0}\dfrac{1-\cos\,x}{x^{2}}=\dfrac{1}{1+1}\\\\\\\boxed{\boxed{\lim\limits_{x\to0}\dfrac{1-\cos\,x}{x^{2}}=\dfrac{1}{2}}}$

Answer 2

\lim_{x->0} \frac{1-cos x}{x^{2}}=\lim_{x->0} \frac{(1-cos x)(1+cos x))}{x^{2}(1+cos x)}=\lim_{x->0} \frac{1-cos^{2} x}{x^{2}(1+cos x)}=\lim_{x->0} \frac{sen^{2}x}{x^{2}(1+cos x)} \newline \lim_{x->0} \frac{sen^2x}{x^{2}}.\frac{1}{(1+cos x)}= 1.\frac{1}{1+cos 0}=\frac{1}{2}