Limit of $1 - \cos(x)$ as $x$ tends to zero.

Question

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 $1$?

Answer 1

Without resorting to continuity:

$$1-\cos x=2\sin^2\dfrac x2.$$

Then taking the limit of $\sin x/x$ for granted,

$$\lim_{x\to0}2\sin^2\frac x2=2\left(\lim_{x\to0}\frac{\sin\dfrac x2}{\dfrac x2}\dfrac x2\right)^2=2\left(\lim_{x\to0}\dfrac x2\right)^2.$$

Answer 2

It is a continuos function in $0$ thus

$$1-\cos(x)\to 1-\cos(0)=0$$

Answer 3

$\cos{(x)}\;is \;a \;{\color{red} {Continuous \;function}}$, So: $$\lim_{x\rightarrow 0}[1-\cos{(x)]}=1-\cos{(0)}=0$$

Answer 4

You can use the l'Hospital rule to get: $$ \lim_{x \to 0} \frac{1-\cos(x)}{x} = \lim_{x \to 0} \frac{\sin(x)}{1} = 0. $$