I have to prove that $\sinh x$ and $\cosh x$ are continuous functions. I have to use the hyperbolic addition formula, and the inequalities:
$|\sinh x| \leq 3|x|, \, |x|<\frac{1}{2}$
$|\cosh x -1| \leq 3|x|, \, |x|<\frac{1}{2}$
I've tried doing an $\varepsilon-\delta$ proof, but I am only able to show it for $|x|<1/2$ or not at all. I believe it's beneficial to first look at $x\in[0,\infty )$ and then use symmetry to extend the argument to all of $R$. Here's what I've tried:
$\cosh x \,\mathrm{is\, continious \,on} \, R_+ \Leftrightarrow \forall x\in R_+ \forall \epsilon >0\exists\delta>0: |\cosh x-\cosh y|<\varepsilon\, \mathrm{ such\, that}\, |x-y|<\delta$
$|\cosh x - \cosh y| = |\cosh x -1 + 1 - \cosh y| = |(\cosh x -1) - (\cosh y -1)|$
$\leq |3|x|-3|y|| = 3||x|-|y||\leq3|x-y|$
And since $|x-y|<\delta$, one can pick a $\delta = \frac{\varepsilon}{3}$. The problem is, this is only valid for $0<x<\frac{1}{2}$ (and thus not on the desired the interval) because of the inequality I used. Also, I haven't used the hyperbolic addition formula.
This thread: Prove cosh(x) and sinh(x) are continuous. deals with the same problem, but my problem must not be solved simply by stating that $\sinh x$ and $\cosh x$ consists of continious functions, as that thread does.
If you have to, the addition formula $\cosh(x+h)=\cosh(x)\cosh(h)+\sinh(x)\sinh(h)$ and $\sinh(x+h)=\cosh(x)\sinh(h)+\sinh(x)\cosh(h)$ gives $$ \begin{aligned} \cosh(x+h)-\cosh(x)&=\cosh(x)[\cosh(h)-1]+\sinh(x)\sinh(h)\\ \sinh(x+h)-\sinh(x)&=\sinh(x)[\cosh(h)-1]+\cosh(x)\sinh(h). \end{aligned} $$ So by triangle inequality, the given bound, and $\cosh(x)+\sinh(x)=e^x$, we have $$ \begin{aligned} \lvert\cosh(x+h)-\cosh(x)\rvert&\leq\cosh(x)\overbrace{[\cosh(h)-1]}^{\leq 3\lvert h\rvert}+\lvert\sinh(x)\overbrace{\sinh(h)}^{\leq 3\lvert h\rvert}\rvert\leq 3 e^x\lvert h\rvert \\ \lvert\sinh(x+h)-\sinh(x)\rvert&=\lvert\sinh(x)\rvert\underbrace{[\cosh(h)-1]}_{\leq 3\lvert h\rvert}+\cosh(x)\underbrace{\lvert\sinh(h)\rvert}_{\leq 3\lvert h\rvert}\leq 3e^x\lvert h\rvert\\ \end{aligned} $$ Hence if we choose $\delta<\min(\frac13\epsilon e^{-x},\frac12)$, we get $\lvert h\rvert<\delta\implies \lvert f(x+h)-f(x)\rvert\leq\epsilon$ for both $f=\cosh$ and $f=\sinh$.