I want to show that, the function of cosine is continuous in $0$. So i have to proof that:
$$\lim_{\theta \to 0}cos(\theta)=1$$
So, i did that:
$$\lim_{\theta \to 0}cos(\theta)=1 \Leftrightarrow (\forall \epsilon>0,\exists \delta>0; 0<|\theta|<\delta \Rightarrow |cos(\theta)-1|<\epsilon )$$
if $|cos(\theta)-1|<\epsilon$, then $(1-\epsilon)< cos(\theta) < (1+\epsilon)$. Which $(\epsilon>0) \Rightarrow 1<\epsilon+1$ and for $(0<\epsilon<2) \Rightarrow 1-\epsilon>-1$.
So let consider $(0<\epsilon^{'}<2) \in \mathbb{R}$, where if $0<\epsilon<2$ then $\epsilon = \epsilon^{'}$, else $\epsilon^{'}<\epsilon$.
$$|cos(\theta)-1|<\epsilon \Leftrightarrow (1-\epsilon) \leq (1-\epsilon^{'}) < cos(\theta) <(1+\epsilon^{'}) \leq (1+ \epsilon) $$
Now we have that $-1<(1-\epsilon^{'})<cos(\theta)\leq1<(1+\epsilon^{'}):$
$$\begin{cases}1-\epsilon^{'}<cos(\theta) \\1 \geq cos(\theta)\end{cases} \Leftrightarrow 2k\pi-cos^{-1}(1-\epsilon^{'})<\theta<2k\pi+cos^{-1}(1-\epsilon^{'})$$
get $k=0$ and $\theta \neq 0$$, 0<|\theta|<cos^{-1}(1-\epsilon^{'})$.
Let $\delta= cos^{-1}(1-\epsilon^{'})$:
$$0< |\theta| <cos^{-1}(1-\epsilon^{'}) \Rightarrow (1-\epsilon^{'})<cos(\theta)<1 \Leftrightarrow (1-\epsilon)\leq(1-\epsilon^{'})<cos(\theta)<(1+\epsilon^{'})\leq(1+\epsilon)$$ Hence: $$0<|\theta|<cos^{-1}(1-\epsilon^{'}) \Rightarrow |cos(\theta)-1|<\epsilon$$.
This is right? and how others ways u can show me?