Can someone help me with this problem.
We have $$x(t)=N \sin (w_{0} t)+\frac{w_0}{w_1}e^{\frac{-t}{T}}\sin (w_{1}t)$$
and $w_1=(1+\frac{\delta_1}{N^2})w_0$ for some $|\delta_1|\leq 1$.
I need to show that $$x(t)=N(1-e^{\frac{-t}{T}})\sin (w_0 t)+ \mathrm{O}(1).$$
I obtained $x(t)=N\sin w_0 t+\frac{N^2}{\delta_1+N^2}e^{-\frac{t}{T}}\sin w_1t$ but I don't know how to continue. I think I need to use the fact that $|\sin(a)-\sin(b)|\leq |a-b|$ but I am not exactly sure how to use it here.
Let's form the difference $$ x(t)-N\left(1-e^{-t/T} \right)\sin(w_0t)=\frac{w_0}{w_1}e^{-t/T}\sin(w_1t)+Ne^{-t/T}\sin(w_0t). $$ Now simply use the fact that $|\sin y|\leq 1$ for every $y\in\mathbb{R}$ to get $$ \left|x(t)-N\left(1-e^{-t/T} \right)\sin(w_0t)\right|\leq\frac{|w_0|}{|w_1|}e^{-t/T}+|N|e^{-t/T}. $$ The rhs is continuous on $[0,+\infty)$ and it tends to $0$ as $t$ tends to $+\infty$. So it is bounded and we get the $O(1)$ we want. Actually, we even have $o(1)$.