Limit the limit

Let $f(x)=1-\cos~x$ and $g(x)=x^2$ . Since, $\large \lim_ {x\rightarrow 0} \dfrac{f(x)}{g(x)}=\dfrac{0}{0}$ , we can use the L'Hospital's Rule

$\large \lim_ {x\rightarrow 0} \dfrac{1-\cos~x}{x^2}$

$\large= \lim_ {x\rightarrow 0} \dfrac{\sin~x}{2x}=\dfrac{0}{0}$ $\implies$ use the L'Hospital's Rule again

$\large=\lim_ {x\rightarrow 0} \dfrac{\cos~x}{2}$

$\large=\dfrac{\cos~0}{2}$

$=$ $\large \boxed{\dfrac{1}{2}}$

Using $\lim_{x \to 0} \frac{\sin x}{x} = 1$ we obtain

$\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \lim_{x \to 0} \frac{(1 - \cos x)(1 + \cos x)}{x^2(1 + \cos x)} = \lim_{x \to 0} \frac{1 - (\cos x)^2}{x^2(1 + \cos x)} = \lim_{x \to 0} \frac{(\sin x)^2}{x^2(1 + \cos x)} = \lim_{x \to 0} (\frac{\sin x}{x})^2 \frac{1}{1+cos x} = 1^2 \frac{1}{1+cos 0} = \frac{1}{2}$