Does the limit of $1 - \cos(x)$ as $x$ tends to zero exist? If yes, what is it? Can it be that the limit of $$ \frac{1-\cos(x)}{x} $$ as $x$ tends to zero is $1$?
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$\begingroup$Without resorting to continuity:
$$1-\cos x=2\sin^2\dfrac x2.$$
Then taking the limit of $\sin x/x$ for granted,
$$\lim_{x\to0}2\sin^2\frac x2=2\left(\lim_{x\to0}\frac{\sin\dfrac x2}{\dfrac x2}\dfrac x2\right)^2=2\left(\lim_{x\to0}\dfrac x2\right)^2.$$
$\endgroup$ 2 $\begingroup$It is a continuos function in $0$ thus
$$1-\cos(x)\to 1-\cos(0)=0$$
$\endgroup$ 4 $\begingroup$$\cos{(x)}\;is \;a \;{\color{red} {Continuous \;function}}$, So: $$\lim_{x\rightarrow 0}[1-\cos{(x)]}=1-\cos{(0)}=0$$
$\endgroup$ 1 $\begingroup$You can use the l'Hospital rule to get: $$ \lim_{x \to 0} \frac{1-\cos(x)}{x} = \lim_{x \to 0} \frac{\sin(x)}{1} = 0. $$
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