Problem with cosh and sinh We consider a parameter  $\theta>0$ 
For all $t>0$ we note:
$$u(t)=\frac{\sinh\left(\frac{t}{2}\cosh(\theta)\right)}{\cosh(\theta)}$$
$$A(t)=\frac{\sqrt{\cosh^2(\theta)u^2(t)+1}-1}{\cosh^2(\theta)}-\Big(\cosh(t/2)-1\Big)=\frac{\cosh\left(\frac{t}{2}\cosh(\theta)\right)-1}{\cosh^2(\theta)}-\Big(\cosh(t/2)-1\Big)$$
$$f(t)=-\ln\Big(1-\frac{2A(t)}{u(t)+\sinh(t)+A(t)}\Big)$$
I want to prove that 
$$f(t)=O(\cosh^2(\theta)t^3)$$ for all $t>0$ such that $$\cosh(\theta)t\le 1-\delta$$
(where $\delta$ has to be precised).
Thanks
 A: To simplify a problem substitute $x=\cosh(\theta)$. Then we have $x\ge 1$, 
 $u(t)=\frac{\sinh\frac{xt}{2}}{x}$, $A(t)=\frac{\cosh\frac{tx}{2}-1}{x^2}-\cosh \frac{t}{2}+1$, and $f(t)=-\ln (1-g(t))$, where $$g(t)=\frac{2A(t)}{u(t)+\sinh t+A(t)}=$$
$$\frac{2\frac{\cosh\frac{tx}{2}-1}{x^2}-2\cosh \frac{t}{2}+2}{\frac{\sinh\frac{xt}{2}}{x}+\sinh t+\frac{\cosh\frac{tx}{2}-1}{x^2}-\cosh \frac{t}{2}+1}=$$
$$\frac{2\cosh\frac{tx}{2}-2-2x^2\cosh \frac{t}{2}+2x^2}{x\sinh\frac{xt}{2}+x^2\sinh t+\cosh\frac{tx}{2}-1-x^2\cosh \frac{t}{2}+x^2}.$$ 
Now given $xt\le 1-\delta$ we want to show that $f(t)=O(x^2t^3)$. Since $x\ge 1$, $t$ is bounded from above, so we investigate the behavior of the function $f(t)$ when $x$ is constant and $t$ tends to zero. In this case for each constant $c$ we have $\sinh ct=ct+O(t^3)$, $\cosh ct=1+\frac {c^2t^2}2+O(t^3)$, so 
$$g(t)=\frac{2+\left(\frac{tx}{2}\right)^2-2-x^2\left(2-\left(\frac{t}{2}\right)^2\right)+2x^2+O(t^3)}{x\frac{xt}{2}+x^2t+1+\frac 12\left(\frac{tx}{2}\right)^2-1-x^2\left(1+\frac 12\left(\frac{t}{2}\right)^2\right) +x^2+O(t^3)}=$$ $$\frac {t^2+O(t^3)}{3t+O(t^3)}=\frac t3+O(t^2).$$
Thus when $t$ tends to $0$ the function $f(t)$ tends to $0$ too, but not as fast as we want. Namely, 
$$f(t)= -\ln (1-g(t))= -\ln\left(1-\frac t3+O(t^2)\right)=\frac t3+o(t)\ne O(x^2t^3).$$
