Find: $$\lim_{n\rightarrow\infty}\frac{\sin\left(x+\frac{1}{n}\right)-\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)}$$ as $x\in\mathbb{R}$ My progress:
$$\lim_{n\rightarrow\infty}\frac{\sin\left(x+\frac{1}{n}\right)-\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)}=\lim_{n\longrightarrow\infty}\frac{\sin\left(x+\frac{1}{n}\right)}{\sin\left(x+\frac{1}{n}\right)}-\frac{\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)}=$$
$$\lim_{n\longrightarrow\infty} \ \ {1}-\frac{\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)}=1-\lim_{n\longrightarrow\infty}\frac{\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)}$$ at this point I got stuck.
I can't evaluate the Taylor series of $\sin(x+\frac{1}{n})$ because $n$ is not fixed
(even if we'll suppose that there exist some $\epsilon>0$ and there exists $ N\in\mathbb{N}:\forall n\geq N$ s.t: $$-\epsilon<\frac{1}{n}<\epsilon$$ it doesn't seem like a formal argument to me)
(I might be very wrong - it's only my intuition).
Also trying to apply L'Hopital's rule for this expression isn't much helpful.